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Help with another code please


Go to solution Solved by JpGemo,

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This code is to change the result in one of my tables but the result won't change

 

This is the form to submit the values to the processing page

<?php 
//connect
include 'connect.php';
//head
include 'head.php';
//colour table
$form = $_POST['selform'];
$sql = "SHOW TABLES FROM $db_name";
$result = mysqli_query($con,$sql);


echo '<form id="form1" name="form1" method="post" action="/BYO/modifyproc.php">';
  echo '<table width="75%" border="0">';
    echo '<tr>';
echo '<td>Table</td>';
echo '<td>';
echo '<select name = "table" id = "table" >';
//list tables 
while ($row = mysqli_fetch_array($result)) {
echo '<option>'.$row[0].'</option>';
}
//formtable 2
    echo '</select>';
echo '</td>';
    echo '</tr>';
    echo '<tr>';
    echo '<td>Column</td>';
    echo '<td><input name="column" type="text" id="column" size="50" value=""/></td>';
    echo '</tr>';
echo '<tr>';
    echo '<td>Result</td>';
    echo '<td><input name="colour" type="text" id="colour" size="50" value=""/></td>';
    echo '</tr>';
    echo '<tr>';
    echo '<td> </td>';
    echo '<td><input type="submit" name="Submit" value="Register"/></td>';
    echo '</tr>';
   echo '</table>';
echo '</form>';
?>

This page receives the form data and update the table

 

<?php 
//connect
include 'connect.php';
//head
include 'head.php';
//recieve posts
$_POST["table"];
$_POST["column"];
$_POST["colour"];
//convert posts
$tablename=$_POST["tablename"];
$column=$_POST["column"];
$colour=$_POST["colour"];
//add to table
mysqli_query($con,"UPDATE $tablename SET $column='$colour'");


echo 'table', $_POST["table"];
echo 'column', $_POST["column"];
echo 'result', $_POST["colour"];
?>

All of the results echo to the page, and the page source doesn't show any incorrect values, but the table isn't updating with the new value

 

Any suggestions would be great :)

 

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https://forums.phpfreaks.com/topic/290987-help-with-another-code-please/
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in your previous thread, it was suggested to use mysqli_error($con) to display query errors. what worked before, can work again.

 

you would be getting a query error at or right after the $tablename portion of the query because you are using a non-existent $_POST variable to provide the value for the $tablename variable. you would also be getting a php undefined index error where you referenced that non-existent $_POST variable.

the $tablename is in the submission

'<select name = "table" id = "table" >';
$_POST["table"];
$tablename=$_POST["tablename"];
mysqli_query($con,"UPDATE $tablename SET $column='$colour'");
echo 'table', $_POST["table"];

and the final echo, echo's the result I submit from the first form.

 

Sorry, I see my error in this code

Edited by JpGemo
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