I am working on Simple Admin Panel,
The method i am working on is to select the data from database and put it into textarea and behind the textarea update button,
when i update the textarea click update to execute query to update the table
but when i click update at the first row for example it execute the third row only even if i clicked the first row update button " picture attached "
<?php
include 'config.php';
echo '<link rel="stylesheet" href="style.css"type="text/css">';
$result = mysql_query("SELECT * FROM English");
while($row = mysql_fetch_array($result))
{
echo "<form action='' method='post'>";
echo "<table>";
echo "<tr>";
echo "<td><textarea rows='1' cols='1' name='txtid' readonly style='overflow:auto;resize:none'>" . $row['ID'] . "</textarea></td>";
echo "<td><textarea rows='4' cols='50' name='txtarea'>" . $row['Content'] . "</textarea></td>";
echo "<td><input type='submit' name='button' value='Update!'/></td>";
echo "</tr>";
}
echo "</table>";
echo "</form>";
if(isset($_POST['button'])){
$textarea =$_POST['txtarea'];
$id = $_POST['txtid'];
$sql = "UPDATE English SET Content='".$textarea."' WHERE ID='".$id."'";
echo $textarea; echo $id;
mysql_query( $sql, $conn );
}
mysql_close($conn);
?>
Example
Well,
Changed the location for closing braces
to be after
echo "</table>";
echo "</form>";
it fixed the problem
Related
I am trying to update a php form that holds a few rows of mysql data. I have a button next to each row and when i click on that I want to update the row. The issue im having below is the ID is only set as the last row. How do i get this to push the ID to the button? So basically no matter what button i press i always get the same ID which is the last one to load.
if($result){
while($row = mysqli_fetch_array($result)){
$id = $row["ID"];
$beername = $row["BeerName"];
$beertype = $row["BeerType"];
$beerpercent = $row["BeerPercent"];
$beerdescription = $row["BeerDescription"];
$nowpouring = $row["NowPouring"] =='0' ? '' : 'checked=\"checked\"';
$glutenreduced = $row["GlutenReduced"] =='0' ? '' : 'checked=\"checked\"';
$beertogo = $row["BeerToGo"] =='0' ? '' : 'checked=\"checked\"';
echo "<form action='' method='POST'>";
echo "<tr><td><h6><input type=\"text\" size=\"5\" name=\"id\" value=\"$id\"></h6></td>";
echo "<td><h6><input type=\"text\" size=\"30\" name=\"BeerName\" value=\"$beername\"></h6></td>";
echo "<td><h6><input type=\"text\" size=\"30\" name=\"BeerType\" value=\"$beertype\"></h6></td>";
echo "<td><h6><textarea size=\"90\" style=\"width:250px;height:150px;\" name=\"BeerDescription\" value=\"\">$beerdescription</textarea></h6></td>";
echo "<td><h6><input type=\"text\" size=\"5\" name=\"Percent\" value=\"$beerpercent\"></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"NowPouring\" value=\"true\" $nowpouring></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"GlutenReduced\" value=\"true\" $glutenreduced></h6></td>";
echo "<td><h6><input type=\"checkbox\" name=\"BeerToGo\" value=\"true\" $beertogo></h6></td>";
#echo "<td><h6> <a href=\". $_SERVER["PHP_SELF"] .?id=".mysql_result($result,$j,'id')."\" onclick=\"\"></h6></td>";
echo "<td><h6> <button name=\"submit\" type=\"submit\" value=\"$id\">Save</button></h6></td>";
echo "</tr>";
echo "</form>";
}
}
if (isset($_POST['submit'])) {
$user = $_POST['submit'];
echo "<p style=\"color:#ffffff\">$id</p>";
#$delet_query = mysqli_query($mysqli, "UPDATE NowPouring SET NowPouring = '1' WHERE ID = '4'") or die(mysql_error());
if ($delet_query) {
echo '<p style="color:#ffffff">Beer with id '.$id.' is updated. To refresh your page, click ' . ' <a href=' . $_SERVER["PHP_SELF"] . ' > here </a></p>';
}
}
?>
The main problem I see here is that the while loop your code has is generating the same name for the inputs...
All of your "<button name=\"submit\" type=\"submit\" value=\"$id\">Save</button>" will have the same name, that's why it always has the last id as value.
Maybe you should try something such as..
<button name=\"$id_submit\" type=\"submit\" value=\"$id\">Save</button>
or if you want you can store it in an array..
<button name=\"submit[]\" type=\"submit\" value=\"$id\">Save</button>
You are seeing this result because the 'name' of each of your inputs is the same, so essentially you have a form with a bunch of elements that have the same names. You need to add a dynamic aspect to each name.
For example, you could update your output to something like this:
echo "<tr><td><h6><input type=\"text\" size=\"5\" name=\"id_$id\" value=\"$id\"></h6></td>";
Where each line adds the current id. Then when you retrieve the form data, you can append the submitted id to the field you want to update.
Have you considered using an AJAX approach so you can submit just the line in question and not have to reload the page and return the whole data set each time?
Make <form> for each submit button. Adding <form> in the while():
if($result){
while($row = mysqli_fetch_array($result)){
echo "<form action='' method='POST'>";
//...
echo "</form>";
}
}
Your form tag is placed at the wrong place.
It should be within:
while($row = mysqli_fetch_array($result)){
$id = $row["ID"];
//....
//....
echo "<form action='' method='POST'>";
echo"<tr>";
echo "<td>" . $id . "</td>";
//....
//....
echo "<td><button type='submit' name='submit'>Save</button></td>";
echo"</tr>";
echo "</form>";
}
I'm trying to make a HTML table as a frontend to a mySQL database. The table displays fine and I can type in the edits I want to make to each row of the table but when I press the submit button the changes aren't actually made. Can anyone see where I'm going wrong?
<?php
include("db.php");
$sql = "SELECT * FROM `artist`";
$result = mysqli_query($conn, $sql);
if (isset($_POST['update'])){
$artID = $_POST['artID'];
$artName = $_POST['artName'];
$key = $_POST['hidden'];
$UpdateQuery = "UPDATE `artist` SET `artID` = '$artID', `artName` = '$artName' WHERE `artist`.`artID` = '$key'";
mysqli_query($conn,$UpdateQuery);
header("Location: {$_SERVER['HTTP_REFERER']}");
exit;
};
echo "<table border='1'>";
echo "<tr>";
echo "<th>ID</th>";
echo "<th>Name</th>";
echo "</tr>";
if ($result->num_rows > 0) {
echo "<form id ='artisttable' action ='getartiststable.php' method ='post'>";
// output data of each row
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" ."<input type='text' name ='artID' value ='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='text' name ='artName' value ='" . $row["artName"] . "' </td>";
echo "<td>" . "<input type = 'hidden' name ='hidden' value='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='submit' name ='update'" . " </td>";
echo "</tr>";
}
echo "</form>";
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
The db.php file simply includes the connection info to the mySQL database and I'm 100% sure there's nothing wrong with it as it retrieves the table correctly it just doesn't update.
You are putting form tag inside tr which is not allowed td are only allowed
so you have to remove that tr from there.
You have to use jquery or you can replace the table with some other grid structure so that it can look the same and the form can be placed there as well
One more suggestion Don't mix the php and html together separate them for the clean code
If you do all these you code will be like this
Your form is being constructed with multiple elements with the same name. When you submit the form it is using the last elements as the values so regardless of the record you want updated the last record is being updated (or throwing an error because of string encapsulation). You should use parameterized queries as well.
So instead of:
if ($result->num_rows > 0) {
echo "<form id ='artisttable' action ='getartiststable.php' method ='post'>";
// output data of each row
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" ."<input type='text' name ='artID' value ='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='text' name ='artName' value ='" . $row["artName"] . "' </td>";
echo "<td>" . "<input type = 'hidden' name ='hidden' value='" . $row['artID'] . "' </td>";
echo "<td>" . "<input type='submit' name ='update'" . " </td>";
echo "</tr>";
}
echo "</form>";
echo "</table>";
Use:
if ($result->num_rows > 0) {
// output data of each row
while($row = mysqli_fetch_array($result)) {?>
<form class='artisttable' action ='getartiststable.php' method ='post'>
<tr>
<td><input type='text' name ='artID' value ='<?php echo $row['artID'];?>' /></td>
<td><input type='text' name ='artName' value ='<?php echo $row["artName"];?>' /></td>
<td><input type = 'hidden' name ='hidden' value='<?php echo $row['artID'];?>' /></td>
<td><input type='submit' name ='update'" . " </td>
</tr>
</form>
<?php } ?>
</table>
So you get a form for each data set. Here's a link on prepared statements with mysqli: http://php.net/manual/en/mysqli.quickstart.prepared-statements.php. You also should update your mark up. Tables for formatting aren't the best approach. Your inputs also weren't closed missing >.
Also this changed artisttable from an id to class because there will be multiples. Update CSS/JS accordingly.
i would like to know how can i make a multiple search criteria with 2 or more textboxes and only one submit button.
my script is:
$sql = "select * from wp_studenti ";
if (isset($_POST['search'])) {
$search_term = mysql_real_escape_string($_POST['search_box']);
$sql .= " WHERE nume= '{$search_term}' ";
}
$query = mysql_query($sql) or die (mysql_error());
echo "<form name ='search_form' method='POST' action='search.php'>";
echo "<center><h3>Cauta:</h3> <input type='text' name='search_box' />";
echo "<input type='submit' name='search' value='Cauta' /></center>";
echo "</form>";
and my results page that shows after search page:
$sql = "select * from wp_studenti ";
if (isset($_POST['search'])) {
$search_term = mysql_real_escape_string($_POST['search_box']);
$sql .= "WHERE nume= '{$search_term}'";
}
echo "<center>\n";
echo "<table border='1'>";
echo "<thead>";
echo "<tr><th>Id</th>";
echo "<th>Nume</th>";
echo "<th>Localitate</th>";
echo "<th>Judet</th>";
echo "<th>Sector Financiar</th>";
echo "<th>Link</th></tr>";
echo "</thead>";
$rst = mysql_query($sql);
while($a_row = mysql_fetch_assoc($rst)) {
echo "<tr>";
echo "<td>"; echo $a_row['id']; echo "</td>";
echo "<td>"; echo $a_row['nume']; echo "</td>";
echo "<td>"; echo $a_row['localitate']; echo "</td>";
echo "<td>"; echo $a_row['judet']; echo "</td>";
echo "<td>"; echo $a_row['sector_financiar']; echo "</td>";
echo "<td>"; echo "<a href='results.php?id={$a_row['id']}'>{$a_row['link']}</a>" ; echo "</td>";echo "</tr>";
echo "</table>";
$sql = "select * from wp_studenti ";
if (isset($_POST['search'])) {
$search_term_by_Cauta = mysql_real_escape_string($_POST['search_box_1']);
$search_term_by_localitate = mysql_real_escape_string($_POST['search_box_2']);
//If you want both search mandatory, use "AND" Operator otherwise use "OR". If you want approximate search use "LIKE" Operator in bellow SQL
$sql .= " WHERE nume= '{$search_term_by_Cauta }' OR localitate = '{$search_term_by_localitate }' ";
}
$query = mysql_query($sql) or die (mysql_error());
echo "<form name ='search_form' method='POST' action='search.php'>";
echo "<center><h3>Cauta:</h3> <input type='text' name='search_box_1' />";
echo "<h3>localitate:</h3> <input type='text' name='search_box_2' />";
echo "<input type='submit' name='search' value='Cauta' /></center>";
echo "</form>";
Well you need another search box:
echo "<center><h3>Cauta:</h3> <input type='text' name='search_box1' /><input type='text' name='search_box2' />";
And you need to use that value in your SQL:
if (isset($_POST['search'])) {
$search_term1 = mysql_real_escape_string($_POST['search_box1']);
$search_term2 = mysql_real_escape_string($_POST['search_box2']);
$sql .= " WHERE nume= '{$search_term1}' OR nume= '{$search_term2}'";
}
But you will have to do some thinking about how the search should work, is it supposed to match exactly one OR the other? If you want the text to contain instead of exactly match, you can use the syntax nume LIKE '%searchword%'
Use mysqli instead of mysql, which is depreciated. By PHP, something like this;
<form method='post'>
<input type='hidden' name='srch' val='1'>
Search Type1: <input type='text' name='s1'>
<br>
Search Type2: <input type='text' name='s2'>
<button>Submit</button>
</form>
<?php
if(isset($_POST['srch']))
{
if(!empty($_POST['s1']))$search = $_POST['s1'];
else if(!empty($_POST['s2']))$search = $_POST['s2'];
else die ('No criteria entered');
rest of your code...
}
?>
Also see functions like mysqli_real_escape for security reasons.
My idea is very simple, I will have a search box and a submit button.
When user key in the keyword and click on the submit button, results will be shown below with an additional button. Now my problem is I have no idea on how to make the button to be located at bottom right of the table populated.
Please consider the below code for my situation:
<input type="text" name="criteriaInput" style="width: 300px;"> <input type="submit" name="submit" value="GO" />
<?php
if (isset($_POST['submit'])) {
if(isset($_POST['inquiryMethod'])){
error_reporting(0);
$sql = 'SELECT
*
FROM
table
WHERE
fullname REGEXP \''.$_POST['criteriaInput'].'\'' ;
$server = mysql_connect("localhost","root", "");
$db = mysql_select_db("mysql",$server);
$query = mysql_query($sql);
echo "<table class=\"striped\">";
echo "<tr class=\"header\">";
echo "<td>Full Name</td>";
echo "<td>ID</td>";
echo "<td>ID Type</td>";
echo "<td>Issuance Country</td>";
echo "<td>Class</td>";
echo "</tr>";
while ($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>".$row[fullname]."</td>";
echo "<td>".$row[id]."</td>";
echo "<td>".$row[id_type]."</td>";
echo "<td>".$row[issuance_country]."</td>";
echo "<td>{$row['class']}</td>";
echo "</tr>";
}
echo "<form method=\"post\" action=\"CIF_InquiryAction.php\">";
echo "<input type=\"submit\" name=\"create\" value=\"Create\" />";
echo "</form>";
echo "</table>";
}else{
echo "Please select one of the criteria!";
}
}
?>
The submit button with value "Create" did successfully created on existence of data, however it's aligned on top left of the table.
Kindly advice Thank you.
You need to put your button into a table row and cell.
echo "<tr>";
echo "<td colspan=\"5\">"
echo "<form method=\"post\" action=\"CIF_InquiryAction.php\">";
echo "<input type=\"submit\" name=\"create\" value=\"Create\" />";
echo "</form>";
echo "</td>"
echo "</tr>";
Also, your form should probably move to be outside your table.
Editing to show input outside of table:
echo "</table>";
echo "<input type=\"submit\" name=\"create\" value=\"Create\" />";
Here is the form code:
<?php
$con=mysqli_connect("localhost","user","pass","db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT ID, NAME FROM b_sonet_group ORDER BY ID DESC");
echo "<form class='form-vertical login-form' action='step-2.php' method='POST'>";
echo "<h4 class='form-title'>Step One: Choose Your Project</h4><div class='control-group'><div class='controls'>";
echo "<select>";
echo "<option value=''>Choose Your Project</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option name='ID' value='" . $row['ID'] . "'>" . $row['NAME'] . "</option>";
}
echo "</select>";
echo "</div></div>";
echo "<div class='form-actions'><button type='submit' name='submit' class='btn green pull-right'>Proceed to Step Two <i class='m-icon-swapright m-icon-white'></i></button></div></form>";
mysqli_close($con);
?>
What do I need to put on page 2 that retrieves the value ID from the form on the previous page and how do I print it so I can check it is the correct ID?
Simple I know but my brain has packed up and gone home.
You need to change your select to
echo "<select name=\"project\">";
On your second page you can get the value with
echo $_POST['project'];
You need to move the name from option to your select. Then echo $_POST['id'];