Php Noob here need help

[theoretical_dreamer]

need help regarding syntax please check my code tnx

$rs4= mysql_query("SELECT IP4 FROM ipadd ORDER BY IP4")or die(mysql_error());
$info4 = (mysql_fetch_array( $rs4 ));
ctr=31;
if (ctr=.$info['IP4'].)
{
print "The available IP is:".$info4['IP4'].;
}
else
{
Print "<b>IP4:</b> ".$info4['IP4']. " <br>";
while($info4 = mysql_fetch_array( $rs4 ))
{

while (ctr==.$info4['IP4'].)
{
ctr++;
if (ctr!=.$info4['IP4'].)
{
print "It works";
}
}

Print "<b>IP4:</b> ".$info4['IP4'] . " <br>";
}
}

[AndyB]

Don't forget to tell us what errors you get and identify the lines in your code. php provides really useful error messages that (with a little practice) lead you to the trouble spots.

Most of your problems stem from mis-use of the . concatenation operator (which is used exclusively for adding strings to make longer strings. Note also the equality check operator is ==. So, for example:
[code]if (ctr=.$info['IP4'].)
{
print "The available IP is:".$info4['IP4'].;[/code]
Should be:
[code]if (ctr==$info['IP4'])
{
print "The available IP is: ".$info4['IP4'];[/code]

[kenrbnsn]

Your souce is filled with syntax errors. For starters:[list][*]The comparision operator for equality is the double equal "==" not the single equal "="[*]The period is the string concatenation operator, code like [code]<?php if (ctr=.$info['IP4'].) ?>[/code] is completely meaningless. What were you trying to write?[/list]Your code written syntactally correct would look something like this: (whether it's logically correct is another question)
[code]<?php
$rs4= mysql_query("SELECT IP4 FROM ipadd ORDER BY IP4")or die(mysql_error());
$ctr=31; // need a $ infront of the variable and since it doesn't change, move it out of the loop
while ($info4 = mysql_fetch_assoc( $rs4 )) { // make this a "while" loop
    if (ctr == $info['IP4'])
       print "The available IP is:".$info4['IP4'];
    else
       print "<b>IP4:</b> ".$info4['IP4']. " <br>";
    while ($ctr == $info4['IP4'])
    {
         $ctr++;
         if ($ctr != $info4['IP4'])
             print "It works";
    }
    Print "<b>IP4:</b> ".$info4['IP4'] . " <br>";
}
?>[/code]

Ken

[theoretical_dreamer]

tnx for the rply that really help

[theoretical_dreamer]

guys i did wat u said but it still didn't work here's my edited code

$rs4 = mysql_query("SELECT IP4 FROM ipadd ORDER BY IP4")or die(mysql_error());
$info4 = (mysql_fetch_array( $rs4 ));
$ctr=31;

if ($ctr=.$info['IP4'].) //if ctr is equal to the 1st value of IP4 field in the database
{
print "The available IP is:".$info4['IP4'].;
}
else
{
Print "<b>IP4:</b> ".$info4['IP4']. " <br>";

while($info4 = mysql_fetch_array( $rs4 ))

{

while ($ctr=.$info4['IP4'].)
{
$ctr++;
if ($ctr!=.$info4['IP4'].)
{
print "It works";
}
}

Print "<b>IP4:</b> ".$info4['IP4'] . " <br>";
}
}

[AndyB]

No, No, No. You didn't do what we said. You didn't post what the errors were. You didn't identify which lines the errors occurred at.

Change this:
[code]if ($ctr=.$info['IP4'].) //if ctr is equal to the 1st value of IP4 field in the database
{
print "The available IP is:".$info4['IP4'].;[/code]
To this (again, carefully please) and try again.
[code]if ($ctr==$info['IP4']); //if ctr is equal to the 1st value of IP4 field in the database
{
print "The available IP is: ".$info4['IP4'];[/code]

[theoretical_dreamer]

oh sorry... i forgot to turned the error report message in apache, thanks bro