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Hello there, i have a table called "category" that haves 2 columns, that are "category" and "category_en"

 

And im echo the list of data that i have, i can make the listins of columns "category appear but not the data of category_en .

Take a look in my code please

 

 


$qry=mysql_query("SELECT * FROM category", $con);
if(!$qry)
{
die("Query Failed: ". mysql_error());
}


<?php
while($row=mysql_fetch_array($qry))
{
echo "<option value='".$row['category_en']."'>".$row['category_en']."</option>";
}
?>
<?php
while($row=mysql_fetch_array($qry))
{
echo "<option value='".$row['category_en']."'>".$row['category_en']."</option>";
}
?>


$qry=mysql_query("SELECT * FROM category", $con);
if(!$qry)
{
die("Query Failed: ". mysql_error());
}


<?php
while($row=mysql_fetch_array($qry))
{
echo "<option value='".$row['category_en']."'>".$row['category_en']."</option>";
}
?>
<?php
while($row=mysql_fetch_array($qry))
{
echo "<option value='".$row['category_en']."'>".$row['category_en']."</option>";
}
?>

 

First off, I wouldn't stop and start the <?php?> tags for one command. Just write it all inside the same tags. Also, make your query its own variable. Just to clean up the code I would do the following:

 

$get = "SELECT * FROM category";
$qry=mysql_query($get, $con);
if(!$qry)
{
die("Query Failed: ". mysql_error());
}
while($row=mysql_fetch_array($qry))
{
$x = $row['category_en'];
echo "<option value='$x'>''$x</option>";
echo "<option value='$x'>'$x'</option>";
}

 

Correct me if I am wrong, but the echo will just echo the x variable. I am not sure why you while looped and fetched the same entry twice. Also, when inside of an echo use the single quotes.

Edited by computermax2328

The second while loop will not output any data as you already reached the end of the resultset in the first loop.

 

You would need a data_seek instruction to return to the beginning again.

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