pingeyeg
asked on
Same record being displayed for each value
Ok, I've got this thing working right, but I am getting the same record for each value displayed on the page.
<?php
$i = 0;
while ($i < 2) {
$i = $i+1;
if ($i==1) {
echo "<tr>";
}
echo "<td width=\"250\">";
?>
<form action="/admin/delete_prov ider.php?C ompany=<?= $strCompanyname ?>" method="post">
<a href="/admin/edit_provider .php?provi derID=<?= $providerID ?>" >
<?= $strCompanyname ?>
</a>
<a href="/admin/confirm_provi der.php?Co mpany=<?= $strCompanyname ?>">
<input type="button" value="Delete" class="btn2">
</a>
</form>
<?php
echo "</td>";
if ($i==3) {
echo "</tr>";
$i=0;
}
}
?>
<?php
$i = 0;
while ($i < 2) {
$i = $i+1;
if ($i==1) {
echo "<tr>";
}
echo "<td width=\"250\">";
?>
<form action="/admin/delete_prov
<a href="/admin/edit_provider
<?= $strCompanyname ?>
</a>
<a href="/admin/confirm_provi
<input type="button" value="Delete" class="btn2">
</a>
</form>
<?php
echo "</td>";
if ($i==3) {
echo "</tr>";
$i=0;
}
}
?>
Where are $strCompanyname and $providerID being set?
ASKER
This is the entire code block:
<?php
if ($_GET['list'] == "list") {
$SQLstr2 = mysql_query("SELECT tblAdspace.providerID, tblAdspace.strCompanyname FROM tblAdspace")
or die(mysql_error());
while($row = mysql_fetch_array($SQLstr2 )){
$strCompanyname = $row['strCompanyname'];
$providerID = $row['providerID'];
}
?>
<table id="admin_list" align="center" border="1">
<?php
$i = 0;
while ($i < 2) {
$i = $i+1;
if ($i==1) {
echo "<tr>";
}
echo "<td width=\"250\">";
?>
<form action="/admin/delete_prov ider.php?C ompany=<?= $strCompanyname ?>" method="post">
<a href="/admin/edit_provider .php?provi derID=<?= $providerID ?>" >
<?= $strCompanyname ?>
</a>
<a href="/admin/confirm_provi der.php?Co mpany=<?= $strCompanyname ?>">
<input type="button" value="Delete" class="btn2">
</a>
</form>
<?php
echo "</td>";
if ($i==3) {
echo "</tr>";
$i=0;
}
}
?>
<?php
if ($_GET['list'] == "list") {
$SQLstr2 = mysql_query("SELECT tblAdspace.providerID, tblAdspace.strCompanyname FROM tblAdspace")
or die(mysql_error());
while($row = mysql_fetch_array($SQLstr2
$strCompanyname = $row['strCompanyname'];
$providerID = $row['providerID'];
}
?>
<table id="admin_list" align="center" border="1">
<?php
$i = 0;
while ($i < 2) {
$i = $i+1;
if ($i==1) {
echo "<tr>";
}
echo "<td width=\"250\">";
?>
<form action="/admin/delete_prov
<a href="/admin/edit_provider
<?= $strCompanyname ?>
</a>
<a href="/admin/confirm_provi
<input type="button" value="Delete" class="btn2">
</a>
</form>
<?php
echo "</td>";
if ($i==3) {
echo "</tr>";
$i=0;
}
}
?>
ASKER CERTIFIED SOLUTION
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As under_dog said, your display portion needs to be inside the while() loop. As it stood before, your while loop was completing, and only the very last record found in the loop was passed on to the display portion of the script.