PHP post always returns the last value

Question

I'm trying to create a simple PHP page that retrieves and shows the information from the DB, so I do this, which works fine:

echo "<form action='#' method='post'>";
if(!empty($result)){
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        echo "\t<tr>\n";
        echo "<td><input type = 'hidden' name = 'value_id' id = 'value_id' value = '" . $row['id'] . "'/></td>";
        echo "<td><input type = 'text' name = 'name_value' id = 'name_value' value = '" . $row['name'] . "'/></td>\n";
        echo "<td><input type = 'text' name = 'addr_value' id = 'addr_value' value = '" . $row['addr'] . "'/></td>\n";
        echo "<td><INPUT type='submit' name = 'modifyItem' value ='Modify'></td>\n";
        echo "<td><INPUT type='submit' name = 'deleteItem' value ='Delete'></td>\n";
        echo "\t</tr>\n";
    }
}
echo "</form>";

Every row has a delete button, which will allow the user to delete each row individually, so I see if the delete button has been clicked and I proceed to remove the row by getting the ID:

if(isset($_POST['deleteItem'])){
    $querydelete = "DELETE FROM address WHERE ID = " . $_POST["value_id"];
    mysql_query($querydelete) or die('Query failed: ' . mysql_error());
}

The problem is that the ID value is always the same after after the post (which is the ID value of the last row). If I show the ID beforehand for each row, it's the correct one, so the problem must be the value it's getting in the post. What am I doing wrong?

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It's because you're including all your `input` nodes within one `form` node, so the last `input` in the DOM with the same `name` value will take precedence. You could use new `form` nodes for each or a bit of Javascript trickery to get your desired output. — esqew, Jun 23 '14 at 14:51

score 3 · Accepted Answer · answered Jun 23 '14 at 14:52

This is because you are creating many fields with the same name in your form. You can create separate form for every record:

if(!empty($result)){
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "<form action='#' method='post'>"
    echo "\t<tr>\n";
    echo "<td><input type = 'hidden' name = 'value_id' id = 'value_id' value = '" . $row['id'] . "'/></td>";
    echo "<td><input type = 'text' name = 'name_value' id = 'name_value' value = '" . $row['name'] . "'/></td>\n";
    echo "<td><input type = 'text' name = 'addr_value' id = 'addr_value' value = '" . $row['addr'] . "'/></td>\n";
    echo "<td><INPUT type='submit' name = 'modifyItem' value ='Modify'></td>\n";
    echo "<td><INPUT type='submit' name = 'deleteItem' value ='Delete'></td>\n";
    echo "\t</tr>\n";
    echo "</form>";
    }
}

score 1 · Answer 2 · answered Jun 23 '14 at 14:51

You have multiple inputs with the name value_id. Only the last occurence of that is transmitted.

What you want is to split into multiple forms. One for each entry. But this would not produce valid html code in your example. There are various ways around that, but this is another question.

score 1 · Answer 3 · answered Jun 23 '14 at 14:52

You're using the SAME field names for all of your field sets. Since you're not using the PHP-specific [] array extension/hint on the names, PHP will only process the LAST field name set submitted.

e.g.

<input type="text" name="foo" value="a" />
<input type="text" name="foo" value="b" />
<input type="text" name="foo" value="c" />

var_dump($_POST);

Array(1) {
   'foo' => 'c'
}

<input type="text" name="foo[]" value="a" />  <--note the [] here
<input type="text" name="foo[]" value="b" />  <--note the [] here
<input type="text" name="foo[]" value="c" />  <--note the [] here

var_dump($_POST);

Array(1) {
   'foo' => array(
       0 => 'a'
       1 => 'b'
       3 => 'c'
   )
}

score 1 · Answer 4 · answered Jun 23 '14 at 14:52

try:

if(!empty($result)){
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "\t<tr>\n";
    echo "<form action='#' method='post'><td><input type = 'hidden' name = 'value_id' id = 'value_id' value = '" . $row['id'] . "'/></td>";
    echo "<td><input type = 'text' name = 'name_value' id = 'name_value' value = '" . $row['name'] . "'/></td>\n";
    echo "<td><input type = 'text' name = 'addr_value' id = 'addr_value' value = '" . $row['addr'] . "'/></td>\n";
    echo "<td><INPUT type='submit' name = 'modifyItem' value ='Modify'></td>\n";
    echo "<td><INPUT type='submit' name = 'deleteItem' value ='Delete'></td>\n";
    echo "\t</form></tr>\n";
    }
}

pavel · Answer 5 · 2014-06-23T19:20:18.260

1

You have one form with more name=value_id elements. When you submit the form, in $_POST is last one (there is the first one, overwrite by second one, etc. to last one).

Use for each row new form, or use array in your names, eg. name="value[]".

edited Jun 23 '14 at 19:20

answered Jun 23 '14 at 14:53

pavel

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score 1 · Answer 6 · answered Jun 23 '14 at 14:57

I have already do the same with JS and PHP :

HTML : [index]

echo "<td> <button name=\"$row['id']\" type=\"clear\" >Clear</button>  </td>";

JavaScript : [index]

<script>
$(document).ready(function(){
     $('[type="clear"]').on('click', function(){ 
         document.location.href = "./php/yourScript.php?param=" + $(this).attr("name");
     })
});
</script>

PHP : [./php/yourScript.php]

<?php
$querydelete = "DELETE FROM address WHERE ID = " . $_GET["param"];
    mysql_query($querydelete)
?>

I hope, that can help you :)

Syam Mohan M P · Answer 7 · 2014-06-23T15:07:08.953

Change your code like this if you want to place all controls in one form.

echo "<form action='#' method='post'>";
if(!empty($result)){
 while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    echo "\t<tr>\n";
    echo "<td><input type = 'hidden' name = 'field[" . $row['id'] . "][value_id]' id = 'value_id-" . $row['id'] . "' value = '" . $row['id'] . "'/></td>";
    echo "<td><input type = 'text' name = 'field[" . $row['id'] . "][name_value]' id = 'name_value-" . $row['id'] . "' value = '" . $row['name'] . "'/></td>\n";
    echo "<td><input type = 'text' name = 'field[" . $row['id'] . "][addr_value]' id = 'addr_value-" . $row['id'] . "' value = '" . $row['addr'] . "'/></td>\n";
    echo "<td><INPUT type='submit' name = 'field[" . $row['id'] . "][modifyItem]' value ='Modify'></td>\n";
    echo "<td><INPUT type='submit' name = 'field[" . $row['id'] . "][deleteItem]' value ='Delete'></td>\n";
    echo "\t</tr>\n";
 }
}
echo "</form>";

adding form tags outside the while loop causes my Delete to always grab the last value. — dcparham, Jul 19 '17 at 13:41

score 1 · Answer 8 · answered Oct 05 '19 at 23:14

To solve this you need to wrap each into separate form...

while(rows_available){
     echo "<form action='#' method='post'>"
     echo "<td><input type = 'hidden' name = 'value_id' id = 'value_id' value = '" . $row['id'] . "'/></td>";
        echo "<td><input type = 'text' name = 'name_value' id = 'name_value' value = '" . $row['name'] . "'/></td>\n";
        echo "<td><input type = 'text' name = 'addr_value' id = 'addr_value' value = '" . $row['addr'] . "'/></td>\n";
        echo "<td><INPUT type='submit' name = 'modifyItem' value ='Modify'></td>\n";
        echo "<td><INPUT type='submit' name = 'deleteItem' value ='Delete'></td>\n";
        echo "\t</tr>\n";
        echo "</form>";
}//end of while loop

PHP post always returns the last value

8 Answers8