MySQL Question??

Kevin_Isley

Hi guys,

Oy! this is a scarey one for me. I hope someone can help me out here.

I need to be able to have listed in a select box data from a MySQL Query.

And it works perfectly on the first time. But I'm guessing that it has something to do with the mysql_fetch_array function. I am guessing that I only can use this once in for loop. Here's what I tried to do.

All help is appreciated. Thanks,

Kevin

CODE

<?php
$count = 15;
for ($i = 0; $count > $i; $i++){

$num = $i;
echo "<select name='listing$num'>";
echo "<option value='' SELECTED>No Label</option>";

while ($InfoNum > $c){
$Info = mysql_fetch_array($Result);
echo "<option value='$Info[listFileId]'>$Info[listFileId] || $Info[lastName]</option>";
$c++;
}

echo "</select><br>";

}

The first select box works no problem but everyone after that doesn't.

But it does display the other select boxes, it's just that all they show is the "No Label" option.

Davidc316

As far as MySQL goes, you want to CONNECT, SELECT then QUERY. The order is very important.

After you've done that bit (all standard stuff), you simply say the following...

while ($row=mysql_fetch_array($result)) {
extract($row);
//then do whatever you want with the database variables
//for the purposes of this demo, I'll simply print a list of results
print "$variable-name-from-thedatabase<br>";
}

You might need to change some variable names, but I hope you get the general idea.

Kevin_Isley

No, I'm not Totally PHP/MySQL dumb.

Maybe I wasn't clear enough. But I found a temporary solution.
The only problem is that the file has gone from a 5kb php file to being a 6MB file after processing.

I need to take data and instead of printing it each time for each select box. I was wondering if this process can be done in JavaScript so that each select box could just make a reference to a variable with all of the data in it, instead of printing all the bunches of <option value=blah>blah</option> for each select box.

Can anyone help with this????

Thanks in advance.

Kevin

Davidc316

I appologise for being so unhelpful but, with respect, I think you are being a little bit unclear about exactly what it is you want to do.

Let me just clarify this...

You have some variables stored on a MySQL database. Correct?

You want to call upon those stored variables and for each and every stored variable you want to have a select option generated on a webpage. Correct?

The reason for you wanting to do this is because you want to create some kind of dynamic online form, with a drop down menu. Correct?

silent

Maybe I can help -

<?php

$stored_result = $result; //save the result first!

while($res = mysql_fetch_array($result)) {
  //do whatever you need to do
}

//do it again, if you need to
$result = $stored_result; //you can use it again!

while($res = mysql_fetch_array($result)) {
//ok?
}

?>

I think this works...

superwormy

Perhaps mysql_data_seek() is what you're after?

Everytime you call mysql_fetch_array() the internal pointer which keeps track of which row you're on is incremented. So if you go through all the rows on the first select box, you need to 'rewind' that pointer back to the beginning before you can loop through all hte records again.

silent

I knew there was a function for that...
hehehe

I didn't know what to look for in the PHP manual though.