I am having a problem calling random number from the database.
Here is the my link:
<form method="post" action="pages/test.php?id=<?php echo "$id[0]";?>&ran=<?php echo "$ran";?>">
<input type="submit" name="Submit" value="Click here" class="button">
</form>
And here is my second page code:
<?php
if (!isset($submit)) {
//database connection here
$id = $_GET['id'];
$ran = $_GET['ran'];
$query2 = "SELECT * FROM table WHERE id='$id' AND ran='$ran'";
$result2 = mysql_query ($query2) or die ('Could not run query: ' . mysql_error());
$info = mysql_fetch_array ($result2);
?>
When I call for example firstname or lastname it woks okay. But $ran is returning empty. Am I missing something.
Whilst I don't totally understand your request, one notable error I can see is below.
Your second page should be like this
<?php
if (isset($_POST)) {
//database connection here
$id = $_POST['id'];
$ran = $_POST['ran'];
$query2 = "SELECT * FROM table WHERE id='$id' AND ran='$ran'";
$result2 = mysql_query ($query2) or die ('Could not run query: ' . mysql_error());
$info = mysql_fetch_array ($result2);
}
?>
Related
I have HTML code with checkbox and submit button as below
<form action="checkboxes.php" method="post">
<input type="checkbox" name="checkbox1" value="Yes">4K</input>
<input type="submit" name="formSubmit" value="Submit" ></input>
</form>
And in my PHP I have a file "config.php" that his function is to connect to my database:
<?php
/* Database connection */
$sDbHost = 'localhost';
$sDbName = 'testowanie';
$sDbUser = 'root';
$sDbPwd = '';
$dbcon = mysqli_connect ($sDbHost, $sDbUser, $sDbPwd, $sDbName);
?>
And a second PHP file:
<?php
include('config.php');
$sqlget = "SELECT * FROM monitory";
$sqldata = mysqli_query($dbcon, $sqlget)or die("Can't connect to the database");
if(isset($_POST['checkbox1']) &&
$_POST['checkbox1'] == 'Yes')
{
while($row = mysqli_fetch_array($sqldata, MYSQLI_ASSOC)) {
echo '.';
echo $row['cena'];
}
}
?>
This all three connected files each others do that if the checkbox is checked this SQL statement are executed: SELECT cena FROM monitory; but I want to execute this statement "SELECT * FROM monitory WHERE cena=1000;
I tried to do this like around 2 hours but I really don't know how to do this.
So You want to choose one of two different queries according to input conditions. Then do it so :-)
<?php
include('config.php');
if (isset($_POST['checkbox1']))
$sqlget = "SELECT * FROM monitory WHERE cena = 1000";
else
$sqlget = "SELECT * FROM monitory";
$sqldata = mysqli_query($dbcon, $sqlget)or die("Can't connect to the database");
while($row = mysqli_fetch_array($sqldata, MYSQLI_ASSOC)) {
echo '.';
echo $row['cena'];
}
I feel like I just need another set of eyes on this. There is of course something in the database to search, however nothing is displayed. Is there something wrong with the syntax or logic. This is all in one file index.php
<form action = "index.php" method = "post">
Search: <input type="text" name="value" placeholder="Is it part of the FWO?"></input>
<input type=submit name = "search" value="Search">
</form>
New Entry
<br>
<p>Search Results</p>
<hr />
<?php
error_reporting(E_ALL);
$title = $_POST['value'];
echo "You have searched: " .$title;
echo "<br>";
$con = mysql_connect("localhost", "user", "pass") or die ('Could not connect, this is the error: ' . mysql_error());
mysql_select_db("db") or die ('Sorry could not access database at this time. This is the error: ' . mysql_error());
$clean = msql_real_escape_string($_GET['value']);
echo "Another test ". $clean;
$run = mysql_query("SELECT * FROM db WHERE name = '$clean'") or die(mysql_error());
if(mysql_num_rows($run) >= 1){
echo "found entry";
while($i = mysql_fetch_array($run)){
echo $i['creator'];
}
}
else {
echo "No entries found";
}
mysql_close($con);
?>
</body>
</html>
Your form is using post method and you are trying get a value by $_GET
instead of this:
$clean = msql_real_escape_string($_GET['value']);
Use this:
$clean = msql_real_escape_string($_POST['value']);
Or
$clean = msql_real_escape_string($title);
To search inside mysql you should use LIKE. and if you want to search anywhere in the string you should encapsulate with %. for example:
$run = mysql_query("SELECT * FROM db WHERE name LIKE '%$clean%'") or die(mysql_error());
for more info: http://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html
I'm trying to update the database after select a few options and submit (using isset). The page display some information after a database query, and then I have to update the database according to the selected option.
If I run exactly the same query that is inside the function "actualizarEstado" but on a third page, then it works. What am I doing wrong? I can't understand. Tried to kill and close the first connection, but I get the same results. Thanks in advice!
<?php
include '00-conexion.php';
$data = extract($_GET);
$sql = "SELECT * FROM inscripciones WHERE nroInscripcion = $sel";
$retval = mysqli_query($conexion, $sql);
$fila = mysqli_fetch_array($retval);
if(isset($_POST['ejecutar'])){
actualizarEstado($sel);
}
function actualizarEstado($sel){
$estado = $_POST['estado'];
$sql2 = "UPDATE inscripciones SET revision1='',
revision2='',
revision3='',
revision4='',
revision5='',
revision6='',
revision7='',
revision8='',
revision9='',
estado='$estado'
WHERE nroInscripcion = $sel";
if (!mysqli_query($conexion, $sql2)) {
die('Error: ' . mysqli_error($conexion));
}
}
?>
<form method="post" action="">
<tr><td><select name="estado" from="estado">
<option value="aceptado">Aceptar</option>
<option value="rechazado">Rechazar</option>
<option value="revision">En revision</option>
</tr/></td>
<tr><td><input type="submit" value="Actualizar" name="ejecutar" onclick="return confirm('¿Estás seguro que deseas?')" /></tr></td>
</form>
I think you have to pass the $connexion variable through the function!
Make it something like this:
function actualizarEstado($myConnection,$sel){
$estado = $_POST['estado'];
$sql2 = "UPDATE inscripciones SET revision1='',
revision2='',
revision3='',
revision4='',
revision5='',
revision6='',
revision7='',
revision8='',
revision9='',
estado='$estado'
WHERE nroInscripcion = $sel";
if (!mysqli_query($myConnection, $sql2)) {
die('Error: ' . mysqli_error($myConnection));
}
}
and call the function like:
actualizarEstado($connexion,$sel);
after I managed to connect my website form to my database, I decided to try to transfer over my files to my work computer.
Initially I only had one error: mysqli_fetch_row() expects parameter 1 to be mysqli_result, boolean given in...
However now I get an extra mysqli_fetch_row() error the same as above but the error is on a different line.
Additionally I also get the error: Undefined index: fill which I never got before. Are there any mistakes in my code? The form still works and can connect to my database.
<center><form action="fill.php" method="post">
Fill
<input type="text" id="fill"" name="fill">
<input type="submit" id ="submit" name="submit" value="Submit here!">
</form></center>
</div>
<?php
$val1 = $_POST['fill'];
$conn = mysqli_connect('localhost', 'root', '')or
die("Could not connect");
mysqli_select_db($conn, 'rfid');
$val2 = "SELECT * FROM card_refill WHERE refill = $val1";
$result1= $conn->query($val2);
$row = mysqli_fetch_row($result1);
$refill1 = $row[2];
$value = "SELECT *FROM card_credit ORDER BY id DESC LIMIT 1:";
$result = $conn->query($value);
$row = mysqli_fetch_row($result);
$refill = $row[2];
$money= $refill+$refill1;
echo $money;
$sql = "UPDATE card_credit SET value = '$money'";
if ($conn->query($sql) === TRUE) {
echo "Success";
}
else {
echo "Warning: " . $sql . "<br>" . $conn->error;
}
mysqli_close($conn);
?>
</body>
</html>
You're getting that error because you use $_POST['fill'] without checking whether it's set first. It will only be set when the form is submitted, not when the form is first displayed. You need to put all the code that processes the form input into:
if (isset($_POST['submit'])) {
...
}
BTW, you can do that entire update in a single query.
UPDATE card_credit AS cc
CROSS JOIN card_refill AS cr
CROSS JOIN (SELECT * FROM card_credit ORDER BY id DESC LIMIT 1) AS cc1
SET cc.value = cr.col2 + cc1.col2
WHERE cr.refill = '$val1'
Like GolezTrol said from his comment. You're mixing object and functional notation.
Although this might not work exactly how you need it to because I don't have all the information. I have written you something I think is close to what you're looking for.
<?php
// Define the below connections via $username = ""; EXTRA....
// This is best done in a separate file.
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$val1 = $_POST['fill'];
$result1 = $conn->query("SELECT * FROM card_refill WHERE refill = '$val1' ");
$result2 = $conn->query("SELECT * FROM card_credit ORDER BY id DESC LIMIT 1:");
$refill1 = array(); // Pass Results1 Into Array
while($row = $result1->fetch_assoc()) {
$refill1[] = $row[2];
}
$refill = array(); // Pass Results2 Into Array
while($row = $result2->fetch_assoc()) {
$refill[] = $row[2];
}
/* Without an example of what data you are getting from your tables you will have to figure out what data you want from the arrays.
$money= $refill+$refill1;
echo "DEBUG: $money";
*/
// This code will not be functional until your populate the $money value.
$sql = "UPDATE card_credit SET value = '$money' ";
if ($conn->query($sql) === TRUE) {
echo nl2br("Record updated successfully"); // DEBUG
print_r(array_values($refill1)); // DEBUG
print_r(array_values($refill)); // DEBUG
echo nl2br("\n"); // DEBUG
} else { // DEBUG
echo "Error updating record: " . $conn->error; // DEBUG
echo nl2br("\n"); // DEBUG
}
$conn->close();
?>
i'm trying to delete a row from database and i`m using the following code. When i submit the page reloads and if i check the db the row is still there. No error, no nothing. What can i do to solve this?
HTML
<div class="delete_row">
Sterge
<form method="post" action="">
*<input type="text" name="id_col" Placeholder="Id-ul coloanei"><br>
<input type="submit" name="submit1" value="Sterge">
</form>
</div>
PHP
$id_stergere=isset($_POST["id_col"]);
$submitcheck2=isset($_POST["submit1"]);
if($submitcheck2 && $id_stergere !==0 ){
$sql = "DELETE FROM evenimente WHERE ID_even=$id_stergere";
$result = query_mysql($sql);
}
if (isset($_POST["id_col"]))
$id_stergere=$_POST["id_col"];
if (isset($_POST["submit1"]))
$submitcheck2=$_POST["submit1"];
//additional check:
// if (!is_numeric($id_stergere)) die('there was a problem');
// if (!$submitcheck2) die('there was a problem');
$sql = "DELETE FROM evenimente WHERE ID_even=".$id_stergere;
$result = mysql_query($sql);
query_mysql?
Use this for your php :
error_reporting(E_ALL ^ E_NOTICE);
if(isset($_POST['submit1'])){
$id_stergere=$_POST["id_col"];
$submitcheck2=$_POST["submit1"];
if($submitcheck2 && $id_stergere){
$sql = "DELETE FROM evenimente WHERE ID_even=$id_stergere";
$result = mysql_query($sql);
}
}
You should be using mysqli instead though.
Use this instead of the $sql = and $result = :
$link = mysqli_connect("localhost","db","password","user") or die("Error " . mysqli_error($link));
$query= "DELETE FROM evenimente WHERE ID_even=$id_stergere";
$result= mysqli_query($link,$query);
To get the numrows use this:
$numrows = $result->num_rows;
For a fetch_array:
while($row = $result->fetch_array()){
$var= $row['field'];
}
UPDATE: Add error_reporting(E_ALL ^ E_NOTICE); to the top of php script.