Multiple Arrays - only 1 working?

[Catfish]

Hello,

I am trying to have a form generated by the DB and then give the user the ability to select yes or no on that form for each entry pulled out of the DB. See the following...

echo "<TR>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><INPUT TYPE='checkbox' NAME='allowed[]' VALUE='yes' $checked>Allow to<br> $caption?</TD>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><INPUT TYPE='checkbox' NAME='allowed[]' VALUE='no' $checked>Disallow to<br> $caption?</TD>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><b>$firm_orgtitle</b><br>$firm_address<br>$firm_city, $firm_state $firm_zip<br>$firm_id</TD>
</TR><INPUT TYPE='hidden' name='firm_id[]' value='$firm_id'>";

then....

foreach($firm_id as $firm_id => $firm_id_insert){
foreach($allowed as $allowed => $allowed_insert){

$sql11d = "INSERT into permissions set case_id='$case_id', firm_id='$firm_id_insert', allow='$allowed_insert'";
$sql_result11d = mysql_query($sql11d,$connection) or die("Invalid query: " . mysql_error());

header("Location:settings.php?tag=success11");
}
}

Well, this not workie. It is putting yes or no into the DB as a seperate row, but the firm_id is the first one from the original query in each new row. Can someone tell me the right way to do this? Thanks

Shawn

[bretticus]

Since all your checked checkboxes on your form are in the allowed[] array, for each firm_id iteration, you will insert all the answers for the whole page.

You might want to use a radio button instead (since this is what they are made for 😉 ) and not use the [] notation in your HTML. Instead just append the $firm_id to your radio inputs.

echo "
<TR> 
    <TD WIDTH='33%' VALIGN='TOP'><span class='main'>
      <INPUT TYPE='radio' NAME=''allowed_$firm_id' VALUE='yes' $checked>
      Allow to<br>
      $caption?</span></TD>
    <TD WIDTH='33%' VALIGN='TOP'><span class='main'>
      <INPUT TYPE='radio' NAME='allowed_$firm_id' VALUE='no' $checked>
      Disallow to<br>
      $caption?</span></TD>
    <TD WIDTH='33%' VALIGN='TOP'><span class='main'><b>$firm_orgtitle</b><br>
      $firm_address<br>
      $firm_city, $firm_state $firm_zip<br>
      $firm_id</span></TD>
  </TR>
  <INPUT TYPE='hidden' name='firm_id[]' value='$firm_id'>
";

Now use this syntax to get your answers

foreach($firm_id as $firm_id_insert)
{ 
	$sql11d = "INSERT into permissions set case_id='".$_POST['case_id'].", firm_id='$firm_id_insert', ";
	$sql11d .= "allow='".$_POST['allowed_'.$firm_id_insert];
	$sql_result11d = mysql_query($sql11d,$connection) or die("Invalid query: " . mysql_error());	
}

header("Location:settings.php?tag=success11");

[Catfish]

bretticus...thanks for your reply, it is much appreciated. I am a little lost though, I see you wrote this...."<INPUT TYPE='radio' NAME=''allowed_$firm_id' VALUE='yes' $checked>" I think the additional (') is incorrect? I took it out because it did not work when it was there.

But more importantly, I am getting "Invalid query: 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 'id_3f778bd444869', allow='' at line 1"

Since I am new to the way this insert statement looks like, I am lost as to where (') and (") should be. Can you guide me in the right direction? Thanks

Shawn

[Catfish]

Does anyone else know the answer to my question? Anyone at all. I really would like to wrap this up and am spending so much time trying to get it, that it is killin me so any help would be appreciated. Thanks

Shawn

[bretticus]

Send me your code (or a sanitized version) or post it here and I'll look at it. You were right, they extra ' was a typo...sorry.

[Catfish]

echo "<TR>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><INPUT TYPE='radio' NAME='allowed$firm_id' VALUE='yes' $checked>Allow to<br> $caption?</TD>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><INPUT TYPE='radio' NAME='allowed$firm_id' VALUE='no' $checked>Disallow to<br> $caption?</TD>
<TD WIDTH='33%' VALIGN='TOP'><span class='main'><b>$firm_orgtitle</b><br>$firm_address<br>$firm_city, $firm_state $firm_zip<br>$firm_id</TD>
</TR><INPUT TYPE='hidden' name='firm_id[]' value='$firm_id'>";
}
echo "</TABLE>
<INPUT TYPE='hidden' name='successtag' value='11b'>
<INPUT TYPE='hidden' name='case_id' value='$case_id'>
<INPUT TYPE='submit' NAME='permissions' VALUE='Update'></FORM>";
}

if ($successtag == "11b") {

foreach($firm_id as $firm_id_insert)
{

$sql11d = "INSERT into permissions set case_id='".$POST['case_id'].", firm_id='$firm_id_insert', ";
$sql11d .= "allow='".$POST['allowed_'.$firm_id_insert];
$sql_result11d = mysql_query($sql11d,$connection) or die("Invalid query: " . mysql_error());

}

header("Location:settings.php?tag=success11");

}

Thanks so much for responding.

Shawn

[bretticus]

you were missing a single quote in your query that you are building right after $_POST['case_id']. The following query should be proper:

$sql11d = "INSERT into permissions set case_id='".$_POST['case_id']."', firm_id='$firm_id_insert', "; 
		$sql11d .= "allow='".$_POST['allowed_'.$firm_id_insert];

[Catfish]

I copied and pasted your SQL statement once again and got....

Invalid query: 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 ''yes' at line 1

Any other ideas?

[Catfish]

this worked!!!

$sql11d = "INSERT into permissions set case_id='".$_POST['case_id']."', firm_id='$firm_id_insert', allow='".$_POST['allowed_'.$firm_id_insert]."'";
$sql_result11d = mysql_query($sql11d,$connection) or die("Invalid query: " . mysql_error());   

Any reason I cannot use that?

[bretticus]

I answer questions the same way I code...much too fast with plenty of typos. I DO SEE that the 'allow' part needs a single quote on the very end. If I were to actually test my code, I'd get the error & fix it in a second 'cause I'm used to this stuff (as it's my profession), however, I sometimes forget that the 'newbies' area is for people just starting out. And when you guys see code, you expect it to work without really understanding it sometimes.

ANYWAYS...looking MORE carefully...looks like your query should be trouble free as you have confirmed.

[Catfish]

No problem bretticus, I just am really confused as to the following...

'".$_POST['case_id']."',

and

$sql11d .=

I just have been writing it like....

INSERT into master set field1='$variable1', etc....

Never seen the '". .'" or the $sql .=

so I am confused and could not even begin to see where things were supposed to go. I am assuming that since you wrote it, it is a better way of doing things, I am trying to understand as to why.

Thanks for your help and it did get me to the working solution!!!

Shawn

[bretticus]

sure, $_POST is a predefined variable for the contents of the post elements. You can use auto_globals=on (which you must have set in php.ini) and not use the notation, but for security reasons (etc.), you should use the super globals.

More Information on Superglobals

Basically, using the $_POST array just ensures that your external variables really are posted by a browser.

the .= is how you concatenate using php (Javascript uses += for example) which basically says, take what's already here in this variable and append this to it. I used it for readibility only to shorted the line length.

Using "something".$anArray['key']."something more" is also concatenation of strings in PHP. Where we have an array, we break the string and add a variable to it. In other words, in PHP you can get away with "some text $variable some more text." But not with "some text $arrayvariable[somekey] some more text." I don't believe you can anyway, but that's bad coding practice either way. You can use {} inside a string optionally and then you don't have to concatenate (ex. "something {$anArray['key']} something more" should work.)

More info on strings

The manual has all this info, you have the link above.

Have fun!