SQL wont display results

[matheussousuke]

Well, the page is not displaying the sql results, perhaps there is something wrong in this line, but I still dont know how to fix:

echo "<tr bgcolor=\"$col\"><td><a class=TN href=\"javascript:popUp('preview.php?uname=$a1[uname]&ename=$ename')\"> $a2[0] $a2[1]</a></td><td> $a1[clname] </td></tr>";
 

Is it the

 $a2[0] $a2[1]

part?




The code:

{
 
    $q2 = "select fname, lname from job_seeker_info where uname = \"$a1[uname]\" ";
 
    $r2 = mysql_query($q2) or die(mysql_error());
 
    $a2 = mysql_fetch_array($r2);
 
 
 
     if($col == "cococo")
 
        { 
 
         $col = "dddddd"; 
 
        }
 
                  else
 
        { 
 
        $col = "cococo"; 
 
        } 
 
 
 
    echo "<tr bgcolor=\"$col\"><td><a class=TN href=\"javascript:popUp('preview.php?uname=$a1[uname]&ename=$ename')\"> $a2[0] $a2[1]</a></td><td> $a1[clname] </td></tr>";
 
}
 

If you want the full code, let me know.

[matheussousuke]

Please, someone, I know it's friday, but...

[zorgi]

Quick look and this is what I see:

$a1[uname] should probably be $a1['uname'] lines 3 & 29
$a1[clname] should probably be $a1['clname'] line 29

[matheussousuke]

I put the quotes, and now I get this:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/sitee/public_html/employers/employers15.php on line 83

[zorgi]

Its probably because of the way you construct your query string. What is on line 83?

[matheussousuke]

lol, on line 83 it's the line I altered.

uname=$a1['uname']&ename=$ename')\"> $a2[0] $a2[1]

[zorgi]

uname=$a1['uname']&ename=$ename'

should be

uname=".$a1['uname']."&ename=$ename'

[matheussousuke]

I did this:

    echo "<tr bgcolor=\"$col\"><td><a class=TN href=\"javascript:popUp('preview.php?uname=".$a1['uname']."&ename=$ename')\"> $a2[0] $a2[1]</a></td><td> $a1[clname] </td></tr>";
 
}

It didnt work and the part .$a1['uname']. changed color, showing the code is wrong. =/

[zorgi]

You might want to have a look at String Operators because you did same mistake on the same line more than once.

[matheussousuke]

I've read, but it's very basic, didn't find the solution for my quotes issue.

[matheussousuke]

What if it's not that line, is this part correct?

{
 
    $q2 = "select fname, lname from job_seeker_info where uname = \"$a1[uname]\" ";
 
    $r2 = mysql_query($q2) or die(mysql_error());
 
    $a2 = mysql_fetch_array($r2);
 
 
 
     if($col == "cococo")
 
        { 
 
         $col = "dddddd"; 
 
        }
 
                  else
 
        { 
 
        $col = "cococo"; 
 
        } 
 

[zorgi]

I agree, it is very basic, but you didn't apply those basic principles to your code. Example:

$a = array("uname" => "matheussousuke", "sname" => "surname");

What you are doing to echo uname element of an array is this:

echo $a[uname];

when you should be doing this:

echo $a['uname'];

You are also doing something like this:

echo $a['uname'] $a['sname'];

And that is exactly what that basic lesson is all about:

echo $a['uname']." ".$a['sname'];

Just apply this to your code and let us know how it went ;)

[matheussousuke]

OK, I'll do it, but forgive my newbie lames if you find some. :P

[omerbutt]

<?php 
 echo "<tr bgcolor=".$col."><td><a class=TN href=\"javascript:popUp('preview.php?uname=".$a1['uname']."&ename=$ename')\"> ".$a2[0].$a2[1]."</a>
 </td><td>". $a1['clname']." </td></tr>";
 

try this replace with you code
regards,
Omer Aslam

[matheussousuke]

I tried but look at the code's color:

[zorgi]

Did you actually try to run that code. It looks like your editor is just highlighting strings in that colour.

[matheussousuke]

of course, here:

[matheussousuke]

As you can see, there is nothing between Nível and << Voltar field.

[zorgi]

Is there suppose to be? Can you set border on your table to 1 or inspect that part of your html with firebug so you can see what is echoed? Also do var_dump on all of your variables to see what they contain.

[matheussousuke]

There should be results bellow "Nível", they are:
Name and Knowledge Level.

Can I use the Chrome Console?

[matheussousuke]

Just as I expected, there is nothing below it, it like the php is not displaying the data:

The HTML part from the loaded page:

<table align="center" width="500" cellspacing="0">
 
    <caption align="center"><b> Candidatos para a vaga N° 1 / posição Pessoas com conhecimento em informática </b><br><font size="2"> Esta oferta foi visualizada 10 vezes e há 6 interessado(s). </font> </caption>
 
    <tbody><tr bgcolor="gray" style="font-family:arial; font-size:11; color:000000; font-weight:bold"><td width="150">Nome </td><td>Nível </td></tr></tbody></table>

[zorgi]

Giving me fragments of code like this doesn't help as error can be somewhere else. I absolutly can not see connection between your HTML and your php.

[matheussousuke]

The full code:

<?
 
// Sistema de Empregos baseado no script Diesel Job Site versão 1.4
 
 
 
include_once "accesscontrol.php";
 
include_once "../configuration.inc.php";
 
 
 
$q1 = "select * from job_aplicants t1, job_careerlevel t2 where t1.aplicant = t2.uname and t1.job_id = \"$_GET[job_id]\" order by t2.clnumber desc";
 
$r1 = mysql_query($q1) or die(mysql_error());
 
 
 
//number of offer rerviews
 
$q22 = "select nv from job_post where job_id = \"$_GET[job_id]\" ";
 
$r22 = mysql_query($q22) or die(mysql_error());
 
$a22 = mysql_fetch_array($r22);
 
 
 
//number of offer aplicants
 
$q33 = "select count(aplicant) from job_aplicants where job_id = \"$_GET[job_id]\" ";
 
$r33 = mysql_query($q33) or die(mysql_error());
 
$a33 = mysql_fetch_array($r33);
 
 
 
 
 
echo "<br><br><br><table align=center width=500 cellspacing=0>
 
    <caption align=center><b> Candidatos para a vaga N° $_GET[job_id] / posição $position </b><br><font size=2> Esta oferta foi visualizada $a22[0] vezes e há $a33[0] interessado(s). </font> </caption>
 
    <tr bgcolor=gray style=\"font-family:arial; font-size:11; color:$s3white; font-weight:bold\"><td width=150>Nome </td><td>Nível </td></tr>";
 
 
 
$col = "cococo";
 
 
 
while ($a1 = mysql_fetch_array($r1))
 
{
 
    $q2 = "select fname, lname from job_seeker_info where uname = \"$a1[uname]\" ";
 
    $r2 = mysql_query($q2) or die(mysql_error());
 
    $a2 = mysql_fetch_array($r2);
 
 
 
     if($col == "cococo")
 
        { 
 
         $col = "dddddd"; 
 
        }
 
                  else
 
        { 
 
        $col = "cococo"; 
 
        } 
 
echo "<tr bgcolor=".$col."><td><a class=TN href=\"javascript:popUp('preview.php?uname=".$a1['uname']."&ename=$ename')\"> ".$a2[0].$a2[1]."</a>
 </td><td>". $a1['clname']." </td></tr>";
}
 
/*
    echo "<tr bgcolor=\"$col\"><td><a class=TN href=\"javascript:popUp('preview.php?uname=$a1[uname]&ename=$ename')\"> $a2[0] $a2[1]</a></td><td> $a1[clname] </td></tr>";
*/
 
echo "</table><center><br><a class=TN href=employers7.php> << voltar</a></center>";
 
 
 
?>
 
<? include_once('../footer.php'); ?>

[zorgi]

can you do var_dump($a1) on line 56

[matheussousuke]

Parse error: syntax error, unexpected T_VARIABLE in /home/balcaoce/public_html/employers/employers15.php on line 57

[zorgi]

what is on lines 56,57,58

[matheussousuke]

I have an idea, I sent you a copy of the file through PM.

[zorgi]

Did you do that var_dump?!

[matheussousuke]

I did, I did it closing the tags, cause before I was receiving errors.

I inserted it in the line 56:

?><? var_dump($a1) ?><?

Got no error, but no results.

[zorgi]

First of all you forgot semicolon. Semicolon signifies the end of a PHP statement so you should get some error.

var_dump($a1);

Can you also do var_dump for $r1 on line 52.

[matheussousuke]

Inserted the semicolon, still the same thing. And at line 52 I inserted ?><? var_dump($r1); ?><? and got this result:
resource(16) of type (mysql result)

[zorgi]

Ok it looks like your while loop is never executed becuse there is nothing to loop through. Check your SQL statement. Echo it out and check its well constructed. I would also use command line or phpMyAdmin to get results directly from database.

Also please go to this page: http://php.net/manual/en/language.types.array.php and read "Why is $foo[bar] wrong?"

[matheussousuke]

So

echo "<tr bgcolor=".$col."><td><a class=TN href=\"javascript:popUp('preview.php?uname=".$a1['uname']."&ename=$ename')\"> ".$a2[0].$a2[1]."</a>
 </td><td>". $a1['clname']." </td></tr>";
}

is correct?

[matheussousuke]

But that's so weird. This line

Esta oferta foi visualizada 10 vezes e há 6 interessado(s).

says the job has been viewed 10 times and that there is 6 interested people in it. if it says there is 6 interested people so that means somehow it is counting from real tables?

[zorgi]

about:

echo "<tr bgcolor=".$col."><td><a class=TN href=\"javascript:popUp('preview.php?uname=".$a1['uname']."&ename=$ename')\"> ".$a2[0].$a2[1]."</a>
 </td><td>". $a1['clname']." </td></tr>";

Whay don't you check it? Simply create test.php and copy that code into it. Provide any values to the variables needed and run it.

[matheussousuke]

I did a test, submitted my own job application, and it worked. :S
I dont know, maybe the system delete old applications.

By the way, now I'm having an attachment issue. I'll work around it, and if I have any problem, I go back and open another topic, but I dont think I'll have.

[matheussousuke]

PS: Thank you very much for staying and helping me during 2 days. :D