PHP won't load MySQL data when searching - php

I am trying to make a search form on my site which prints all possible things that have come out of the database. However, when I search for something, it always gives me back that it can't find anything. How do I fix this?
PHP code:
$host = 'localhost';
$user = '111042';
$password = 'jcbvrjd8';
$db_name = '111042';
$search = $_POST["search"];
if(isset($search)) {
$db = mysqli_connect($host, $user, $password, $db_name);
$wild_search = "%".$search."%";
$findname = "SELECT `name`,`surname`
FROM `Account`
WHERE `name` LIKE '".$wild_search."'
OR `surname` LIKE '".$wild_search."'
OR CONCAT(`name`, `surname`) LIKE '".$wild_search."'
OR CONCAT(`name`, `surnameprefix`, `surname`) LIKE '".$wild_search."';";
$query = mysqli_query($db,$findname);
$results = mysqli_fetch_all($query);
if($result) {
echo "<div id='searchresult'>\n";
echo "<h1>People Found:</h1>\n";
echo "<table id='searchresult'>\n";
foreach($result as $rowno => $row) {
echo "<tr class='searchtablerow'>\n";
echo "<td>".$row['name'].", ".$row['surname']."</td>\n";
echo "</tr>\n";
}
echo "</table>\n";
} else {
echo "<div id='searchresult'><h1>People Found:</h1>\n";
echo "<p>No one was found...</p>\n";
echo "</div>\n";
}
} else {
}
HTML form:
<div id="searchform">
<h1>Search friends:</h1>
<form name="searchform" method="post" action ="searchlink.php">
<input type="text" name="search" id="search" autofocus placeholder="e.g. John Smith..."></input> <br>
<input type="submit" name="submitsearch" value="Search" id="searchbutton"></input>
</form>
</div>
Thank you guys for helping me.
PS. Please do not roast me, I'm just not as advanced as you are.

$query = mysqli_query($db,$findname);
$results = mysqli_fetch_all($query);
if ($results) {
...
"s" missing at end of result. (or delete the s in $results = mysqli_fetch_all($query);)

Related

how to update form data with submit button in a while loop

I am trying to figure out mysqli (I am just a starting scripter). I created the following script to grab 3 different values from my database. And it prints it on the screen in different textareas and input fields.
What I want to be able to do is when I press the update button that it'll update the records in the database for the form where the button is attached to.
Can anyone give me some tips on how to achieve something like that?
<?php
$sqlserver = <SQLSERVER>;
$sqluser = <SQLUSER>;
$sqlpassword = <SQLPASSWORD>;
$sqldatabase = <SQLDATABASE>;
$mysqli = new mysqli($sqlserver, $sqluser, $sqlpassword, $sqldatabase);
$loggedinuserid= "5";
$standaardtekstlabel = $mysqli->query("SELECT standaardtekst_label FROM Standaardteksten WHERE standaardtekst_account_pID='".$loggedinuserid."'");
$standaardtekstnl = $mysqli->query("SELECT standaardtekst_tekst FROM Standaardteksten WHERE standaardtekst_account_pID='".$loggedinuserid."'");
$standaardteksten = $mysqli->query("SELECT standaardtekst_tekst_en FROM Standaardteksten WHERE standaardtekst_account_pID='".$loggedinuserid."'");
while ($NL_Tekst = mysqli_fetch_row($standaardtekstnl))
{
$label_Tekst = mysqli_fetch_row($standaardtekstlabel);
$EN_Tekst = mysqli_fetch_row($standaardteksten);
print '<form action="" method="POST">
<input type="text" name="standaardtekst_label" value=' . $label_Tekst[0] . '>
<textarea name="standaardtekst_tekst">' . $NL_Tekst[0] . '</textarea>
<textarea name="standaardtekst_tekst_en">' . $EN_Tekst[0] . '</textarea>
<input type="submit" value="update">
</form>';
}
?>
First of all, there is absolutely 0.0 reason why you're using 3 queries for the information you're trying to get. You can simply have: $standaardtekst = $mysqli->query("SELECT standaardtekst_label,standaardtekst_tekst,standaardtekst_en FROM Standaardteksten WHERE standaardtekst_account_pID='".$loggedinuserid."'");
Now regarding your question that is now probably obsolete:
Make the names of the input like this: standaardtekst_tekst[] saving it in an array.
You also need to have a unique(auto increment) key in your database like: id and put it in every form. You can even use the value of this field in the name like this: standaardtekst_tekst[$id].
You could edit your code a bit to look something like this:
<?php
$sqlserver = <SQLSERVER>;
$sqluser = <SQLUSER>;
$sqlpassword = <SQLPASSWORD>;
$sqldatabase = <SQLDATABASE>;
$mysqli = new mysqli($sqlserver, $sqluser, $sqlpassword, $sqldatabase);
$loggedinuserid= "5";
$q = $mysqli->query("SELECT standaardtekst_id, standaardtekst_label,
standaardtekst_tekst, standaard_tekst_tekst_en
FROM Standaardteksten
WHERE standaardtekst_account_pID='".$loggedinuserid."'");
while ($NL_Tekst = mysqli_fetch_row($standaardtekstnl))
{
$row = mysqli_fetch_row($q);
?>
<form action="" method="POST">
<input type="text" name="formData[<?= $row['id']; ?>][standaardtekst_label]" value="<?= $row['standaardtekst_label']; ?>">
<textarea name="formData[<?= $row['id']; ?>][standaardtekst_tekst]"><?= $row['standaardtekst_tekst']; ?></textarea>
<textarea name="formData[<?= $row['id']; ?>][standaardtekst_tekst_en]"><?= $row['standaardtekst_tekst_en']; ?></textarea>
<input type="submit" value="update">
</form>
<?php
}
?>
What I've done:
Made your 3 queries into a single query
Gave each form a unique id
Cleaned up the code a bit
This enables you to do the following:
if ($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['formData'])) {
foreach ($_POST['formData'] as $id => $value) {
$stmt = $mysqli->query("UPDATE standaardtekst SET standaardtekst_label='".$value['standaadtekst_label']."', standaardtekst_tekst='".$value['standaardtekst_tekst']."', standaardtekst_tekst_en='".$value['standaardtekst_tekst_en']."' WHERE standaardtekst_id='".$id."'");
}
}
Thx for the help everyone. Everything is working right now.
This is the script i used to get it to work :-)
<?php
$sqlserver = <SQLSERVER>;
$sqluser = <SQLUSER>;
$sqlpassword = <SQLPASSWORD>;
$sqldatabase = <SQLDATABASE>;
$mysqli = new mysqli($sqlserver, $sqluser, $sqlpassword, $sqldatabase);
$loggedinuserid= "5";
$result = $mysqli->query("SELECT * FROM Standaardteksten WHERE standaardtekst_account_pID='".$loggedinuserid."'");
$row_s = $result->fetch_assoc();
do{
print '<form action="" method="POST">
<input type="text" name="standaardtekst_label" value=' . $row_s['standaardtekst_label'] . '>
<textarea name="standaardtekst_tekst">' . $row_s['standaardtekst_tekst'] . '</textarea>
<textarea name="standaardtekst_tekst_en">' . $row_s['standaardtekst_tekst_en'] . '</textarea>
<input type="text" name="standaardtekst_ID" value="'. $row_s['standaardtekst_ID'] .'"/>
<input type="submit" value="update">
</form>';
} while($row_s = $result->fetch_assoc());
if ($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['standaardtekst_ID']))
{
$updatesql= sprintf("UPDATE Standaardteksten SET standaardtekst_label='%s', standaardtekst_tekst='%s', standaardtekst_tekst_en='%s' WHERE standaardtekst_ID='%s'",
$_POST[standaardtekst_label],
$_POST[standaardtekst_tekst],
$_POST[standaardtekst_tekst_en],
$_POST[standaardtekst_ID]
);
$mysqli->query($updatesql);
echo "Het volgende wordt aangepast: <br />", "Label:", $_POST[standaardtekst_label], "<br />" , "NL tekst:", $_POST[standaardtekst_tekst], "<br />" , "EN tekst:", $_POST[standaardtekst_tekst_en];
echo "<meta http-equiv='refresh' content='1;url=/form.php'>";
}
?>

looking for something like SQL LIKE Operator

i am looking for same code to search things in my table colmon.
i dont want its search something "like" the input search.
i want it to search the exactly Value in the table.
if user search "the walking dead season 3 episode 1"
its search exactly "the walking dead season 3 episode 1" in the "moviename" row..
and dont show me the all results look like thats..
my code now is :
<?php
function connect($host,$username,$database,$password){
$to_connect = mysql_connect($host,$username,$password) or die ("UNFinded ".$username. " DB !");
$db = mysql_select_db($database, $to_connect) or die (mysql_error());
return $db;
}
connect("localhost","idevice2_ariel","idevice2_ariel","ariel123456");
if (!isset($_POST['submit_form'])) {
echo '<form name="search_form1" method="POST" action="search.php">
<input type="text" name="search_name" value="search movie name" />
<input type="submit" value="Submit" name="submit_form" />
</form>';
} else {
$search_name = $_POST['search_name'];
$query = mysql_query ("SELECT * FROM members where `moviename` = $search_name");
$count = mysql_num_rows($query);
while($row = mysql_fetch_array($query)) {
$fname = $row['moviename'];
$lname = $row['links'];
print ''.$row['moviename'].'<br />';
}
}
?>
thanks you! ariel.
if (!isset($_POST['submit_form'])) {
echo '<form name="search_form1" method="POST" action="search.php">
<input type="text" name="search_name" value="search movie name" />
<input type="submit" value="Submit" name="submit_form" />
</form>';
} else {
$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'world');
$search_name = $_POST['search_name'];
$query = $mysqli->prepare("Select * from members where moviename = ? OR links = ?");
$query->bindParam('ss',$search_name,$search_name);
$query->execute();
$query->bind_result($link,$movieName);
while($query->fetch()) {
echo ''.$movieName.'<br />';
}
$query->close();
}
?>
Consider using prepared statements.

updating a record in database not working

again I'm trying to study php mysql and it seems that I tried everything thing to figure the problem out.. but it seems as a beginner codes in the internet are not helping.. I really can't update the records in the database.
<html>
<body>
<?php
$db = mysql_connect("localhost", "root");
mysql_select_db("dbtry",$db);
$id = isset($_GET['id']) ? $_GET['id'] : null;
$submit = isset($_POST['submit']);
if ($id) {
if ($submit) {
$result = mysql_query("select * from employees where id = " . mysql_real_escape_string($_GET['id']) );
$row = mysql_num_rows($result);
if ($myrow != 0) {
mysql_query ("UPDATE employees SET firstname='$first',lastname='$last',address='$address',position='$position' WHERE id = '$id'");
}
echo "Thank you! Information updated.\n";
} else {
// query the DB
$result = mysql_query("SELECT * FROM `employees` WHERE `id` = " . mysql_real_escape_string($_GET['id']), $db);
$myrow = mysql_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']?>">
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
First name:<input type="Text" name="first" value="<?php echo $myrow["firstname"] ?>"><br>
Last name:<input type="Text" name="last" value="<?php echo $myrow["lastname"] ?>"><br>
Address:<input type="Text" name="address" value="<?php echo $myrow["address"]
?>"><br>
Position:<input type="Text" name="position" value="<?php echo $myrow["position"]
?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?php
}
} else {
// display list of employees
$result = mysql_query("SELECT * FROM employees",$db);
while ($myrow = mysql_fetch_array($result)) {
printf("%s %s<br>\n", $_SERVER['PHP_SELF'], $myrow["id"],
$myrow["firstname"], $myrow["lastname"]);
}
}
?>
</body>
</html>
There are two things potentially causing you a problem: firstly, the values you are trying to set are variables which have not been defined. I'm assuming the begginers code you found assumed you had register globals enabled, you really don't want to do this!
The second problem, is that if you do have register globals enabled, the data isn't being sanitized, so a quotation mark could send the update awry.
Try this instead:
$first = mysql_real_escape_string( $_POST['first'] );
$last = mysql_real_escape_string( $_POST['last'] );
$address= mysql_real_escape_string( $_POST['address'] );
$position = mysql_real_escape_string( $_POST['position'] );
mysql_query ("UPDATE employees SET firstname='$first',lastname='$last',address='$address',position='$position' WHERE id = '$id'");
This should at least get you up and running. I'd strongly advise that you use either the MySQLi library, or PHP PDO, and think about using prepared statements for added security.
mysql_query("UPDATE `employees` SET `firstname`='".$first."', `lastname`='".$last."',
`address`='".$address."', `position`='".$position."' WHERE `id` = '".$id".' ; ", $db) or
die(mysql_error());
I think the problem may lie in your connection to the database. The third parameter of the mysql_connect function is a password. Therefore this:
$db = mysql_connect("localhost", "root");
should be:
$db = mysql_connect("localhost", "root", "yourPassword");
It would also help a lot if you posted what type of error you are getting.
You need to differentiate post and get. Follow the working example below. It will sort you out :D
<html>
<body>
<?php
$db = mysql_connect("localhost", "root","");
mysql_select_db("test",$db);
if($_SERVER['REQUEST_METHOD']=='POST')
{
//SUBMIT FORM
$id=isset($_POST['id'])?$_POST['id']:0;
if ($id) {
$result = mysql_query("select * from parameter where id = " . mysql_real_escape_string($id) );
$rows = mysql_num_rows($result);
if ($rows != 0) {
mysql_query ("UPDATE parameter SET name='".$_POST['name']."',value='".$_POST['value']."' WHERE id = '".$id."'");
echo "Thank you! Information updated.\n";
}
}
}
if($_SERVER['REQUEST_METHOD']=='GET')
{
//SELECT WHERE ID=GER VAR AND DISPLAY
$id = isset($_GET['id']) ? $_GET['id'] :0;//
if ($id) {
// query the DB
$result = mysql_query("SELECT * FROM parameter WHERE `id` = " . mysql_real_escape_string($_GET['id']), $db);
$myrow = mysql_fetch_array($result);
?>
<form method="post" action="<?php echo $_SERVER['PHP_SELF']?>">
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
First name:<input type="Text" name="name" value="<?php echo $myrow["name"] ?>"><br>
Last name:<input type="Text" name="value" value="<?php echo $myrow["value"] ?>"><br>
<input type="Submit" name="submit" value="Enter information">
</form>
<?php
}
else {
// display list of employees
$result = mysql_query("SELECT * FROM parameter",$db);
while ($myrow = mysql_fetch_array($result)) {
echo "<a href='".$_SERVER['PHP_SELF']."?id=".$myrow['id']."'>".$myrow['name'].": ".$myrow['value']."</a><br>";
}
}
}
?>
</body>
</html>
Usually when I run into this problem, it's because auto commit is off and I forgot to tell the connection explicitly to commit.
EDIT: Have you tried this: How can I implement commit/rollback for MySQL in PHP?? Depending on your settings, InnoDB can be set to auto commit off, which means you need to tell MySQL explicitly to commit updates after your done.

PHP IF ELSE - If seems to be ignored?

i'm a bit of PHP noob, so sorry if this is a daft question, but I just can't figure this out by myself so any help would be greatly appreciated!
I am trying to create a modify page for an events web application. It seems to be working except when I try to validate the final if/else statement.
This statement returns the value of $row[0] and checks if its == NULL. If NULL, it should return an echo 'this event does not exist' to the user, if there is a value, it presents the user with a matrix of text boxes that they can change the data in.
Currently it works fine for the else statement when there is data found, but doesnt recognise the original if when there is no data. Coupled with that, the footer at the bottom of the page disappears!
Here is the main body of the code, i have highlighted the problem area. I understand that there is probably a more effective and efficient way of doing it all, but please keep it simple as I'm still learning. Thanks again. Dan
<div class="round">
<div id="main" class="round">
<span class="bold">MODIFY EVENT</span><br /><br />
On this page you can modify or delete the events that are stored on the database.<br />
Be aware that you cannot undo the DELETE function.<br />
<br />
Find Event:<br />
<table>
<form name="form1" id="form1" method="post" action="
<?php echo $_SERVER["PHP_SELF"]; ?>" >
<tr><th>By Date:</td><td><input type="text" name="modifyDate"
id="modifyDate" value="dd/mm/yy" /></td></tr>
<tr><th>By Name:</td><td><input type="text" name="modifyName" id="modifyName"
value="" /></td></tr>
<tr><th>Find All:</th><td><input type="checkbox" name="modifyAll"
id="modifyAll" /><td></tr>
<tr><td></td><td><input type="submit" name="submit" value="Search" /></td></tr>
</form>
</table>
<?PHP
if(!isset($_POST['modify'])){
if(isset($_POST['submit'])){
$moddate = $_POST['modifyDate'];
If($moddate == "dd/mm/yy"){
$date = "";
}
else{
$newDate = str_replace("/",".",$moddate);
$wholeDate = $newDate;
$dateArray = explode('.', $wholeDate);
$date = mktime(0,0,0,$dateArray[1],$dateArray[0],$dateArray[2]);
}
$name = $_POST['modifyName'];
$all = $_POST['modifyAll'];
$host = "localhost";
$user = "user";
$password = "password";
$db = "database";
$con = mysql_connect($host,$user,$password) or die('Could not connect to Server');
$dbc = mysql_select_db($db, $con) or die('Could not connect to Database');
if($all != 'on'){
$q = "SELECT * FROM events WHERE date = '$date' || title = '$name' ";
}
else{
$q = "SELECT * FROM events";
}
$result = mysql_query($q);
$row = mysql_fetch_array($result) or die(mysql_error());
//THIS IS THE PROBLEM HERE!!!!
if($row[0]==NULL){
echo 'This event does not exist';
}
else{
?>
<form name="form1" id="form1" method="post" action="
<?phpecho $_SERVER['PHP_SELF']; ?>" >
<?PHP
$result = mysql_query($q) or die(mysql_error());
while ($row = mysql_fetch_array($result)){
$initialDate = date('d/m/y', $row['date']);
$ID = $row['ID'];
echo '<input type="text" name="inputEmail" id="inputEmail" value="'.$initialDate.'" />';
echo '<input type="checkbox" value=$ID name="toModify[]" style = "visibility: hidden;" /><br /><br />';
}
echo '<input type="submit" name="modify" value="Modify" />
<br /><br />';
}
}
}
else{
//modify database and return echo to user
}
?>
</div>
<div id="clear"></div>
</div>
</div>
</div>
<div id="footer" class="round shadow"></div>
</body>
</html>
mysql_fetch_array($result) returns false if there are no rows, so it's possible that even if there is nothing in the result, it's still returning a false and your if($row[0] == null) is evaluating to false, also.
In other words, you should be doing a more robust test on the return results from your query to catch fringe cases.
As others have mentioned or implied, here are some things you could / should be doing:
test for rows returned, not values in rows
test for existence of variable, not content of variable
check the database for errors returned
Is the field in the table it's pulling $row[0] from set to NOT NULL? If it is, it will never be a null value. Try instead something like (if empty($row[0]))
You could check the number of result rows to see if there are any events:
$result = mysql_query($q);
$numrows = mysql_num_rows ($result);
if($numrows === 0){
...

Deleting Multiple Records using Checkboxes in PHP

I am having an issue where I need to be able to delete multiple records using checkboxes.
Here is the code that I currently have.
<?php
$host = "localhost";
$user = "root";
$pass = "";
$dbName = "ticket_history";
$table_name = "ticket_history";
################ Connect to the Database and SELECT DATA ####################################
$conn = mysql_connect($host, $user, $pass) or die ("Unable to connect");
mysql_select_db($dbName);
$query = "SELECT Date,Ticket_Number,Description,Result FROM $table_name";
$result = mysql_query($query);
$count=mysql_num_rows($result);
#############################################################################################
?>
<HTML>
<HEAD>
<TITLE></TITLE>
</HEAD>
<BODY>
<table width=50%>
<form method="post" action="insert_ticket.php">
<table width border='0'>
<tr><td> Date:<input type="text" name="date"/></td>
<td>Ticket #:<input type="text" name="ticket"/></td></tr>
<table>
<tr><td>Description:<TEXTAREA COLS=50 name="description"></TEXTAREA></td></tr>
<tr><td> Result :<TEXTAREA COLS=50 name="result"></TEXTAREA></td></tr>
<tr><td><input type="submit" name="submit" value="Add"/></td></tr>
</table>
</table>
</form>
<form method="post" action="delete_ticket.php">
<input type="submit" name="delete" value="Delete"/>
</form>
</table>
<?php
print "<table width=80% border=1>\n";
$cols = 0;
while ($get_info = mysql_fetch_assoc($result)){
$id = $get_info->id;
if($cols == 0)
{
$cols = 1;
print "<tr>";
print "<th>Select</th>";
foreach($get_info as $col => $value)
{
print "<th>$col</th>";
}
print "<tr>\n";
}
print "<tr>\n";
print "<td><input type='checkbox' name='selected[]' id='checkbox[]' value=$id></td>";
foreach ($get_info as $field)
print "\t<td align='center'><font face=arial size=1/>$field</font></td>\n";
print "</tr>\n";
}
print "</table>\n";
mysql_close();
?>
<!------------------------------------------------------------!>
</BODY>
</HTML>
Delete.php
<?php
$host = "localhost";
$user = "root";
$pass = "";
$dbName = "ticket_history";
$table_name = "ticket_history";
################ Connect to the Database and SELECT DATA ####################################
$conn = mysql_connect($host, $user, $pass) or die ("Unable to connect");
mysql_select_db($dbName);
$query = "SELECT Date,Ticket_Number,Description,Result FROM $table_name";
$result = mysql_query($query);
$count=mysql_num_rows($result);
#####################################
if($_POST['delete']) {
$checkbox = $_POST['selected'];
$countCheck = count($_POST['selected']);
for($i=0;$i<$countCheck;$i++) {
$del_id = $checkbox[$i];
$sql = "DELETE FROM ticket_history WHERE Auto = $del_id";
$result = mysql_query($sql);
}
}
?>
I just want to be able to delete rows checked. How would I go about doing this effectively and efficiently?
Thank you in advance.
The simple answer to your question would be to use:
$sql = sprintf('DELETE FROM ticket_history WHERE Auto IN ()',
implode(',', $checkbox));
However as people will jump in and tell you, you are vulnerable to SQL injection. You should never trust user input. You are deleting using an ID, which I'm assuming must be an integer.
Using something like this will validate that:
$ids = array();
foreach($_POST['selected'] as $selected) {
if (ctype_digit($selected)) {
$ids[] = $selected;
}
else {
// If one is invalid, I would assume nothing can be trusted
// Depends how you want to handle the error.
die('Invalid input');
}
}
$sql = sprintf('DELETE FROM ticket_history WHERE Auto IN (%s)',
implode(',', $ids));
Other issues:
You seem to be using id's, but have not selected that field in your initial query.
$query = "SELECT Date,Ticket_Number,Description,Result FROM $table_name";
Then you reference:
$id = $get_info->id;
Check the HTML output is actually what you expect.
In your delete query, you are referencing the field Auto. Is that your ID field?
And lastly, there no checking if the user has permission to do so. If this is a public site anyone can delete from that table.
Example of using two submit buttons within one form:
<?php
if (isset($_POST['create'])) {
echo "Create!";
}
elseif (isset($_POST['delete'])) {
echo "Delete!";
}
?>
<html>
<form method="post">
<input type="submit" name="create" value="Create"/>
<input type="submit" name="delete" value="Delete"/>
</form>
</html>

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