PHP while loop (mysql_fetch_array) only displaying last record from table

Question

This is my code:

$result = mysql_query ($query) or die (mysql_error());
$row = mysql_fetch_assoc($result);

while($row = mysql_fetch_array( $result )) {

$contactID=$row['contactID'];
$contactFamily =$row ['contactFamily'];
$Contact=$row ['Contact'];
$contactOwner =$row ['contactOwner'];
$Impact=$row ['Impact'];
$Probability =$row ['Probability'];



if ($row ['Impact']=="5" && $row['Probability']=="4")
{
    $grid1=  $row ['contactFamily']." ".$row ['Contact']." .$row['contactOwner'];
}

?>
<br/>
<?php

if ($row ['Impact']=="3" && $row['Probability']=="4")
{
    $grid2=  $row ['contactFamily']." ".$row ['Contact']." .$row['contactOwner'];
}
?>
<br/>
<?

}

The problem is that it is only displaying the last record in the table when I echo $grid1 or $grid2 rather than all of them.

Please can anyone help?

they're not arrays, so when you assign new values to them, they overwrite the last value — Jeff Hawthorne
– Jeff Hawthorne, Commented Mar 21, 2013 at 18:40

John Conde · Accepted Answer · 2013-03-21 18:40:22Z

0

$grid1 and $grid2 need to be an arrays. Otherwise you're just overwriting your value each time you go through the loop.

$grid1 = array();
$grid2 = array();
while($row = mysql_fetch_array( $result )) {

$grid1[] =  $row ['contactFamily']." ".$row ['Contact']." ".$row['contactOwner'];

$grid2[] =  $row ['contactFamily']." ".$row ['Contact']." ".$row['contactOwner'];

answered Mar 21, 2013 at 18:40

John Conde

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SteveP · Accepted Answer · 2013-03-21 18:41:47Z

0

You should echo $grid1 and $grid2 inside the while loop as each time the loop runs, they are overwritten by the next value.

answered Mar 21, 2013 at 18:41

SteveP

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Chris L · Accepted Answer · 2013-03-21 18:42:19Z

0

Each time you iterate over the results set you are re-assigning $grid1 and $grid2 with the string you are creating, which is why only the last result is being displayed. What you want to do is use the concatenating assignment operator .= e.g.

$grid1 .= $row['contactFamily']." ".$row['Contact']." ".$row['contactOwner'];

answered Mar 21, 2013 at 18:42

Chris L

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This has fixed the problem fine but now I am getting undefined variable notices for $grid1 and $grid2.

user2179827
– user2179827

03/21/2013 19:46:54
Commented Mar 21, 2013 at 19:46

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Farid Rn · Accepted Answer · 2013-03-21 18:49:19Z

You are filling a variable multiple times but it only keeps the last value, that's the reason you always see the last value. I've changed your code using arrays:

<?php
$result = mysql_query($query) or die (mysql_error());
$row = mysql_fetch_assoc($result);
$i = 0
while($row = mysql_fetch_array($result)) {
    $data[$i]['contactID'] = $row['contactID'];
    $data[$i]['contactFamily'] = $row['contactFamily'];
    $data[$i]['Contact'] =$row['Contact'];
    $data[$i]['contactOwner'] = $row['contactOwner'];
    $data[$i]['Impact'] = $row['Impact'];
    $data[$i]['Probability'] = $row['Probability'];

    if ($row['Impact'] == "5" && $row['Probability'] =="4") {
        $grid1[$i] = $row['contactFamily'] . " " . $row ['Contact'] . " " . $row['contactOwner'];
    }
    if ($row['Impact']=="3" && $row['Probability']=="4") {
        $grid2[$i] = $row['contactFamily'] . " " . $row ['Contact'] . " " . $row['contactOwner'];
    }
    $i++;
}

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