Unable to save record in php table

Question

Sophia_1 0 Junior Poster in Training

11 Years Ago

Hi,tried to update the improvement_plan table with the records below but unable to do so, as the records are not updated. When i click submit button, no records are saved. Please advise? Thanks.

<?php
error_reporting(E_ALL ^ E_NOTICE);
mysql_connect("localhost","user","");
mysql_select_db("p1");
$table="improvement_plan";


    if(isset($_POST['submit'])){


  //  echo $_SESSION['Picid'];  
   echo $_SESSION['Progressid'];    


       foreach($_POST['Ipid'] as $value) {



        $Item2 = $_POST['Item2' . $value];
        $Business2 = $_POST['Business2' . $value];
        $Rootcause2 = $_POST['Rootcause2' . $value];
        $Progressid = $_SESSION['Progressid'. $value];      

        mysql_query("UPDATE $table SET Item2='$Item2', Business2='$Business2', Rootcause2='$Rootcause2',Progressid = '$Progressid' where Progressid='".$_SESSION["Progressid"]."'");

        echo "Record ".$value." successfully updated.<br />";
    } 


     } 


 echo "<form action='#' method='post'>


           <table border='1'>


        <tr>
<th>Ipid</th>;
<th>Item</th>;
<th>Business Enablers/ISO Clause</th>;
<th>Root Cause</th>;


            </tr>";



    $Ipid=0;
    if (isset($_GET['Ipid'])) {
    $_SESSION['Ipid']=$_GET['Ipid'];    
    $Ipid=$_SESSION['Ipid'];  
    echo $Ipid;
    }
$sql = "SELECT * FROM $table where Ipid='$Ipid'";
$results = mysql_query($sql);
while($row = mysql_fetch_assoc($results))


    {



        $Ipid = $row["Ipid"];
        $Item2 = $row["Item2"];
        $Business2 = $row["Business2"];
        $Rootcause2 = $row["Rootcause2"];
        $Progressid = $_SESSION["Progressid"];
echo "<tr>


             <td><input type='text' name='Ipid' value='$Ipid' readonly></td>


         <td><input type='text' name='Item2' value='$Item2'></td>
         <td><input type='text' name='Business2' value='$Business2'></td>
         <td><input type='text' name='Rootcause2' value='$Rootcause2'></td>    
         <td><input type='text' name='Progressid' value='$Progressid' readonly></td>   
      </tr>";


     }


echo "</table><input type='submit' name='submit' value='Update'></form>";
?>

php

Edited 11 Years Ago by Member #120589 because: reformatted code

2 Contributors
6 Replies
238 Views
2 Days Discussion Span
Latest Post 11 Years Ago Latest Post by Sophia_1

Dani AI

Generated 7 Months Ago

Two things are most likely blocking updates here: session state and how the form inputs are named/processed.

You are reading/writing $_SESSION['Progressid'] but there’s no session_start() at the top of the script, so that value will be empty and your WHERE clause won’t match anything. Also the code loops as if Ipid (and related fields) are arrays, but the form inputs are named as single fields — PHP won’t create the array you expect. Use a quick debug (for example temporarily add var_dump($_POST); exit;) to see exactly what is posted.

Fix approach (minimal, practical):

Add session_start(); before any output.
Render inputs as row-indexed arrays (so each row posts its values keyed by the primary key).
Use mysqli or PDO prepared statements (not mysql_*) and check errors/affected rows.

Example patterns (form naming + server-side update using mysqli prepared statements):

<!-- form row (server-side rendering) -->
<input type="text" name="rows[<?= $Ipid ?>][Item2]" value="<?= htmlspecialchars($Item2) ?>">
<input type="text" name="rows[<?= $Ipid ?>][Business2]" value="<?= htmlspecialchars($Business2) ?>">
<input type="text" name="rows[<?= $Ipid ?>][Rootcause2]" value="<?= htmlspecialchars($Rootcause2) ?>">
<input type="hidden" name="rows[<?= $Ipid ?>][Progressid]" value="<?= htmlspecialchars($Progressid) ?>">

<?php
session_start();
$mysqli = new mysqli('localhost','user','pass','p1');
$stmt = $mysqli->prepare(
  "UPDATE improvement_plan SET Item2=?, Business2=?, Rootcause2=?, Progressid=? WHERE Ipid=?"
);
foreach ($_POST['rows'] as $ipid => $r) {
  $stmt->bind_param('ssssi', $r['Item2'], $r['Business2'], $r['Rootcause2'], $r['Progressid'], $ipid);
  if (! $stmt->execute()) {
    error_log("Update failed for Ipid=$ipid: ".$stmt->error);
  }
}
$stmt->close();
?>

Checklist before re-testing:

Confirm session_start() and that $_SESSION values are set.
Use var_dump($_POST) to verify the array structure.
Update WHERE to use the table primary key (Ipid) unless you truly intend to match Progressid.
Log or display SQL errors during development (but hide them in production).
Follow ’s advice: migrate off mysql_* and avoid raw concatenation—use prepared statements to prevent SQL injection.

This addresses the common causes seen in the thread and should make the update path deterministic and debuggable.

All 6 Replies

Member #120589

11 Years Ago

Any reason why you're using defunct code (mysql_*)?

Member #120589

11 Years Ago

Look at the mysql_* functions in the php manual and apply error checking.

Example:

$result = mysql_query('SELECT * WHERE 1=1');
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

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Sophia_1 0 Junior Poster in Training · Answer 1 · 2015-02-10T08:43:18+00:00

Hi, have change the update statement as below but still unable to update the records. Please advise. Thanks.

<?php
error_reporting(E_ALL ^ E_NOTICE);
mysql_connect("localhost","user","");
mysql_select_db("p1");
$table="improvement_plan";


        if(isset($_POST['submit'])){


  //  echo $_SESSION['Picid'];  


       echo $_SESSION['Progressid'];    


       foreach($_POST['Ipid'] as $value) {
mysql_query("UPDATE improvement_plan set Item2='" . $_POST["Item2"] . "', Business2='" . $_POST["Business2"] . "', 
Rootcause2='" . $_POST["Rootcause2"] . "',
Progressid='" . $_POST["Progressid"] . "' WHERE Progressid='" . $_POST["Progressid"] . "'");
$message = "Record Modified Successfully";


        } 


     } 
 echo "<form action='#' method='post'>


               <table border='1'>


        <tr>
<th>Ipid</th>;
<th>Item</th>;
<th>Business Enablers/ISO Clause</th>;
<th>Root Cause</th>;


                </tr>";


    $Ipid=0;
    if (isset($_GET['Ipid'])) {
    $_SESSION['Ipid']=$_GET['Ipid'];    
    $Ipid=$_SESSION['Ipid'];  
    echo $Ipid;
    }
$sql = "SELECT * FROM $table where Ipid='$Ipid'";
$results = mysql_query($sql);
while($row = mysql_fetch_assoc($results))


        {


        $Ipid = $row["Ipid"];
        $Item2 = $row["Item2"];
        $Business2 = $row["Business2"];
        $Rootcause2 = $row["Rootcause2"];
        $Progressid = $_SESSION["Progressid"];
echo "<tr>


                 <td><input type='text' name='Ipid' value='$Ipid' readonly></td>


         <td><input type='text' name='Item2' value='$Item2'></td>
         <td><input type='text' name='Business2' value='$Business2'></td>
         <td><input type='text' name='Rootcause2' value='$Rootcause2'></td>    
         <td><input type='text' name='Progressid' value='$Progressid' readonly></td>   
      </tr>";
     }
echo "</table><input type='submit' name='submit' value='Update'></form>";
?>

Member #120589 · Answer 2 · 2015-02-10T11:50:29+00:00

OK a few things.

1) You are using deprecated code: mysql_* functions instead of PDO or mysqli. I believe I mentioned this in my first post.

2) You are inserting POST data directly into your query - this is very dangerous thing to do as you are open to SQL injection. If you don't know what that is, I suggest that you read up on it, as this is a major security hole. Using PDO or mysqli perpared statements and binding parameters/values can obviate this issue. If you insist on using deprecated code - please don't - but if you do, at least use mysql_real_escape_string() as a bare minimum.

3) Avoid mixing up your code and markup. Separate them as much as possible. You can keep php code in separate files and use include/require or at a bare minimum place all php above the doctype declaration. Then only use trivial php in the markup, like echo, for/while, if etc. Mixing everything up makes code maintenance a nightmare and also confuses contributors.

4) Ipid - difficult to know what's going on with all this looping.

An example of a safer method of using mysql here:

$queryTemplate = "UPDATE improvement_plan SET Item2='%s', Business2='%s', Rootcause2='%s' WHERE Progressid='%s'";

foreach($_POST['Ipid'] as $value) {

    $i2 = mysql_real_escape_string($_POST["Item2"]);
    $b2 = mysql_real_escape_string($_POST["Business2"]);
    $r2 = mysql_real_escape_string($_POST["Rootcause2"]);
    $id = mysql_real_escape_string($_POST["Progressid"]);

    mysql_query(sprintf($queryTemplate, $i2, $b2, $r2, $id));
    //check for success before you tell everybody that it was successful!!
    $message = "Record Modified Successfully";
}

Sophia_1 0 Junior Poster in Training · Answer 3 · 2015-02-11T02:22:45+00:00

Hi, thanks for your reply. Have tried to use the coding above but still can't update the records. Please kindly advise. Thanks a lot.

<?php
error_reporting(E_ALL ^ E_NOTICE);
mysql_connect("localhost","user","");
mysql_select_db("p1");
$table="improvement_plan";

    if(isset($_POST['submit'])){
    echo $_SESSION['Progressid'];    
  $queryTemplate = "UPDATE improvement_plan SET Item2='%s', Business2='%s', Rootcause2='%s' WHERE Progressid='%s'";


foreach($_POST['Ipid'] as $value) {


$i2 = mysql_real_escape_string($_POST["Item2"]);
$b2 = mysql_real_escape_string($_POST["Business2"]);
$r2 = mysql_real_escape_string($_POST["Rootcause2"]);
$id = mysql_real_escape_string($_POST["Progressid"]);

mysql_query(sprintf($queryTemplate, $i2, $b2, $r2, $id));
//check for success before you tell everybody that it was successful!!
$message = "Record Modified Successfully";


} 


 }

echo "<form action='#' method='post'>

       <table border='1'>


    <tr>

<th>Ipid</th>;
<th>Item</th>;
<th>Business Enablers/ISO Clause</th>;
<th>Root Cause</th>;
</tr>";

    $ip=0;


if (isset($_GET['Ipid'])) {
$_SESSION['Ipid']=$_GET['Ipid'];    
$ip=$_SESSION['Ipid'];  
echo $ip;
}

$sql = "SELECT * FROM $table where Ipid='$ip'";
$results = mysql_query($sql);
while($row = mysql_fetch_assoc($results))

        {


    $ip = $row["Ipid"];
    $i2 = $row["Item2"];
    $b2 = $row["Business2"];
    $r2 = $row["Rootcause2"];
    $id = $_SESSION["Progressid"];

echo "<tr>

                 <td><input type='text' name='Ipid' value='$ip' readonly></td>


     <td><input type='text' name='Item2' value='$i2'></td>
     <td><input type='text' name='Business2' value='$b2'></td>
     <td><input type='text' name='Rootcause2' value='$r2'></td>    
     <td><input type='text' name='Progressid' value='$id' readonly></td>   
  </tr>";
 }

echo "</table><input type='submit' name='submit' value='Update'></form>";
?>

Sophia_1 0 Junior Poster in Training · Answer 4 · 2015-02-12T01:38:57+00:00

Hi, have tried to apply error checking but still can't update the improvement plan table. Appreciate for your advise. Thanks alot.

$sql = "SELECT * FROM $table where Ipid='$ip'";
$results = mysql_query($sql);

if (!$results) {


die('Invalid query: ' . mysql_error());


}

Unable to save record in php table

Dani AI

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