[RESOLVED] Two arrays help!

kpowning

Ok I have the following arrays being submitted by my form:

I HOPE THIS MAKES SINCE!

$getgirlname = $POST['girlname']; which is a textbox with name="girlname[]"
$getgirlthumb = $POST['girlthumb']; which is a hidden input with a value = "imagenameofuser" name="girlthumb[]"

So I'm trying to run a query that will will go through these arrays but only run depending on how many $girlthumbs[] which are pictures are being posted. If there are (3) thumbnails being posted then I need to get all three thumbnail name from $getgirlthumb and all (3) $getgirlname and run the query (3) times to update my mysql.

If i use foreach i can get their arrays.

foreach ($getgirlname as $key => $value){
$test = "girlname = $value";
}
foreach ($getgirlthumb as $key2 => $value2){
$test2 = "WHERE thumb = $value2";
}

mysql_query("UPDATE user SET $test, $test2"); // update it (3) times if three images with different names

Weedpacket

One thing that you'll need to do is name all the checkboxes so that you know which thumb goes with which name. I.e., girlname[0], girlname[1], ... girlthumb[42]...

What you could further do is name them
girl[0][name], girl[1][name], ... girl[42][thumb].

Then you'll have one array, $POST['girl'], where each element contains either a name, or a thumb, or both - depending on whether the corresponding checkboxes are checked. print_r() it to see exactly what it contains.

foreach($_POST['girl'] as $girl)
{
    if(isset($girl['name']) $test = "girlname=".$girl['name']; // Modulo escaping etc.
    if(isset($girl['thumb']) $test2 = " where thumb=".$girl['thumb']; // ditto
   // do stuff with $test and $test2.
}

kpowning

Weedpacket thank you so much for the help - ONE MORE THING THOUGH PLEASE

I have the following now and mysql update gives me an error near 'Where'
any ideas?

I did like you said for the following:

input type text > name="girl[0][name]"
hidden input > name="girl[0][thumb]"
radio button > name ="girl[0][makemain]"

all of these are surrounded by a while statement so they are in a loop in the actual form. if the user has 3 thumb nails then 3 girl names will appear and 3 radio buttons will appear, and 3 thumbnails appear, but they can only choose one radio button not all 3. (JUST TO CLEAR UP WHAT I'M DOING)

foreach ($_POST['girl'] as $girl){
if (isset($girl['name'])) $query1 = "SET girlname = '".$girl['name']."'";
if (isset($girl['makemain'])) $query2 = ", mainimage = '".$girl['makemain']."'";
if (isset($girl['thumb'])) $query3 = " WHERE thumb = '".$girl['thumb']."'";
$updategirls = "UPDATE images $query1 $query2 $query3";
$Result1 = mysql_query($updategirls, $erotic) or die(mysql_error());
}

Weedpacket

What would be clearer is if I could see the offending query and the full error message MySQL is producing.

But I will note that you'll have a seriously broken query if the name is missing.

Or if the girl's name is "O'hara". (I wonder if Bobby Tables has a sister?)

bradgrafelman

As Weedpacket pointed out (in a roundabout way :p), it appears as though you're vulnerable to SQL injection. User-supplied data should never be placed directly into a SQL query! Instead, sanitize it with a function such as [man]mysql_real_escape_string/man.

EDIT: He also noted that if the name isn't set, your query will also be broken. Since the name seems to be an important part, you could resolve this by either a) checking to see if the name is set (and possibly non-empty, if you wish) and displaying an error message if it is not, or b) move the 'SET' into the query variable so that it is always present.

kpowning

Here is the error displayed on the page.

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE thumb = 'thumb__20071231203975096047792a8f9060c.jpg'' at

doesnt show anything else. The query In the form is:

       <table border="0" align="center">
                  <tr> 
                    <? $i = 0; while($get = mysql_fetch_array($imagepreview)){?>
                    <td><div align="center"><a href="javascript:popImage('http://<? echo $siteaddress; ?>/customerimages/<? echo $get['photo']; ?>','<? echo $_SESSION['username'];?>')"><img src="/customerimages/thumb/<? echo $get['thumb']; ?>" border="1" alt="<? echo $querymeminfo['username']; ?>" title="<? echo $_SESSION['username']; ?>"></a><br>
                        <a href="?action=del&&img=<? echo $get['thumb']; ?>"><font size="1" face="Geneva, Arial, Helvetica, sans-serif">(Delete)</font></a><br>
                    <? if ($_SESSION['multiplegirls'] == 'Yes'){ ?>
					    <font color="#FFFFCC" size="1" face="Geneva, Arial, Helvetica, sans-serif"><u>Anonymous 
                        Name:</u></font><br>
						<input type="hidden" name="girl[<? $i ?>][thumb]" id="girlthumb" value="<? echo $get['thumb']; ?>">
                        <input name="girl[<? $i ?>][name]" type="text" id="girlname" size="15" value="<? echo $_POST['girlname']; ?>">
                        <br>
					<? }// end agency check ?>
                        <input type="radio" name="girl[<? $i ?>][makemain]" id="makemain" value="Yes">
                        <font color="#FFFFCC" size="2" face="Geneva, Arial, Helvetica, sans-serif">Default 
                        Image?</font> </div></td>
                    <? $i++; } ?>
                  </tr>
                </table>

bradgrafelman

There is no SQL query in the code you provided; echo out the query string that's failing and paste it here for us to see.

kpowning

It's already posted above in the foreach() loop. = mysql_query

bradgrafelman

That's the PHP code for the query, not the actual query itself. In order to see what was sent to MySQL, you need to echo out the query variable.

kpowning

But one last thing please. I have the following i had to chage $i to $s because I already have a loop using $i.

                    <? $s = 0; while($get = mysql_fetch_array($imagepreview)){?>
                    <td><div align="center"><a href="javascript:popImage('http://<? echo $siteaddress; ?>/customerimages/<? echo $get['photo']; ?>','<? echo $_SESSION['username'];?>')"><img src="/customerimages/thumb/<? echo $get['thumb']; ?>" border="1" alt="<? echo $querymeminfo['username']; ?>" title="<? echo $_SESSION['username']; ?>"></a><br>
                        <a href="?action=del&&img=<? echo $get['thumb']; ?>"><font size="1" face="Geneva, Arial, Helvetica, sans-serif">(Delete)</font></a><br>
                    <? if ($_SESSION['multiplegirls'] == 'Yes'){ ?>
					    <font color="#FFFFCC" size="1" face="Geneva, Arial, Helvetica, sans-serif"><u>Anonymous 
                        Name:</u></font><br>
	                    <input name="girl[<? echo $s; ?>][name]" type="text" id="girlname" size="15" value="<? echo $get['girlname']; ?>">
                        <br>
					<? } ?>
				     	<input type="hidden" name="girl[<? echo $s; ?>][thumb]" id="girlthumb" value="<? echo $get['thumb']; ?>">
                        <input type="radio" name="girl[<? echo $s; ?>][makemain]" id="girlmakemain">
                        <font color="#FFFFCC" size="2" face="Geneva, Arial, Helvetica, sans-serif">Default 
                        Image?</font> </div></td>
                    <? $s++; } ?>

How can I make it so the radio field is only allowed to be selected once no matter how many are shown?

Right now if the user has 3 images in the database then 3 radio fields are shown and they can select all 3. I just want them to only be able to select one, but still show all 3 radio fields. THANKS FOR ALL THE HELP GUYS. This has been a true learning project that is a little advanced for me.

kpowning

my last post double posted here weird.

Weedpacket

If you want the radio buttons to work as a single group they must all have the same name (which makes sense: only one of them will ever be submitted). Contrariwise, every id attribute value in a page has to be unique (by definition).

So your radio buttons apply to the whole group, not the individual entries. So take them out of the array, give them all the same name, and give them values so that you can tell which one succeeded.

<?php
$imagecount = mysql_num_rows($imagepreview);
for($s=0; $s<$imagecount; $s++)
{
	$get = mysql_fetch_array($imagepreview);
?>
<td>
<div align="center">
<a href="javascript:popImage('http://<?php echo $siteaddress ?>/customerimages/<?php echo $get['photo'] ?>','<?php echo $_SESSION['username'] ?>')">
<img src="/customerimages/thumb/<?php echo $get['thumb'] ?>" border="1" alt="<?php echo $querymeminfo['username'] ?>" title="<?php echo $_SESSION['username'] ?>" />
</a><br />
<a href="?action=del&&img=<?php echo $get['thumb'] ?>" class="deletelink">(Delete)</a><br />
<?php if($_SESSION['multiplegirls']=='Yes') { ?>
<span class="anonymous">Anonymous Name:</span><br />
<input name="girl[<? echo $s; ?>][name]" type="text" size="15" value="<? echo $get['girlname']; ?>" /><br />
<?php } ?>
<input type="hidden" name="girl[<?php echo $s ?>][thumb]" value="<?php echo $get['thumb'] ?>">
<input type="radio" name="makemain" value="<?php echo $s?>">
<span class="default">Default Image?</span>
</div>
</td>
<?php } ?>

Oh, and is it really necessary for that hidden field to be there - can't you just get that from the database later? "<?php" is preferred over "<?" for starting PHP blocks; and the <font> tag has been deprecated since last century.

kpowning

Thank you, so im not going to use the foreach($_POST['girl'] as $girl) loop anymore? Im not sure how to manage this now. I'm lost. Sorry. Weedpacket do you mind trying to explain a little bit more for me or show me an example.

kpowning

Ok here is my new code:

// START POST OF FORM //
if ($_POST['finish']){

foreach ($_POST['girl'] as $girl){

if (isset($girl['name'])) { 
$update1 = "SET girlname ='".$girl['name']."'";
if (isset($girl['thumb'])) $update2 = " WHERE thumb = '".$girl['thumb']."'";
$updategirls = "UPDATE images $update1 $update2";
//echo $updategirls;
$Result1 = mysql_query($updategirls, $erotic) or die(mysql_error());
}
}

echo $_POST['makemain'];

if (isset($_POST['makemain'])){
$updatedefault = "UPDATE images SET mainimage = '".$_POST['makemain']."' where thumb = '".$girl['thumb']."'";
echo $updatedefault;
} else {
$updatedefault = "UPDATE images SET mainimage = '' WHERE thumb = '".$girl['thumb']."'";
echo $updatedefault;
}
$Result2 = mysql_query($updatedefault, $erotic) or die(mysql_error());

HERE IS MY HTML:

<? $s = 0; while($get = mysql_fetch_array($imagepreview)){?> 
                    <td><div align="center"><a href="javascript:popImage('http://<? echo $siteaddress; ?>/customerimages/<? echo $get['photo']; ?>','<? echo $_SESSION['username'];?>')"><img src="/customerimages/thumb/<? echo $get['thumb']; ?>" border="1" alt="<? echo $querymeminfo['username']; ?>" title="<? echo $_SESSION['username']; ?>"></a><br> 
                        <a href="?action=del&&img=<? echo $get['thumb']; ?>"><font size="1" face="Geneva, Arial, Helvetica, sans-serif">(Delete)</font></a><br> 
                    <? if ($_SESSION['multiplegirls'] == 'Yes'){ ?> 
                        <font color="#FFFFCC" size="1" face="Geneva, Arial, Helvetica, sans-serif"><u>Anonymous 
                        Name:</u></font><br> 
                        <input name="girl[<? echo $s; ?>][name]" type="text" size="15" value="<? echo $get['girlname']; ?>"> 
                        <br> 
                    <? } ?> 
                         <input type="hidden" name="girl[<? echo $s; ?>][thumb]" value="<? echo $get['thumb']; ?>"> 
                        <input type="radio" name="makemain" value="<?php echo $s; ?>"> 
                        <font color="#FFFFCC" size="2" face="Geneva, Arial, Helvetica, sans-serif">Default 
                        Image?</font> </div></td> 
                    <? $s++; } ?>

How can I make sure that the 'makemain' is clicked for that certain image? So basically I have a certain number of thumbnails being shown from the database depending on how many a user has uploaded. This script shows them including a name text box to allow them to name each image accordingly. But this also allows them to select which one out of all the images they want as the default - main image to be used throughout the site.

kpowning

Weedpacket - I Appreciate your help and I do understand that I want to use no font information thats no problem, I just wasnt sure if your code follows exactly what I was looking for. If so, I will definitely integrate it. I think I have finally explained clearer what the end result needs to happen.