mysql/php help

imarockstar · October 1, 2008

I have a DB with a crap load of data .. however some data fields are empty and when I want to display all the rows .. but if a certain data field is blank I do not want to display the header ... basically so theres not an empty spot on the page ...

my code for not displaying info if mysql returns no rows ..

if (mysql_num_rows($result) < 1)
{
  echo "You have no resumes for this job posting. ";
}

HOWVER ... what if I have variables .... data1 data2 data3 ...

data1 has info stored ...

data2 has NO INFO

data3 has info stored ..

in my code above how would i not display just data2 ?

make since .. kuz i just confused myself ...

F1Fan · October 1, 2008

It would help to see your code, but I would say with an IF statement.

volatileboy · October 1, 2008

I didnt quite understand what you were propperly saying but something like this would only output something that wasn't empty

if(empty($row->data1)) {

echo $row->data1;

}

volatileboy · October 1, 2008

Correction:

if(!empty($row->data1)) {

echo $row->data1;

}

asmith · October 1, 2008

When you fetch a data , it is an array like this :

array(0 => data1, 1 => data2, 3 => data3)

so you can use foreach to findout if they are empty or not.

while ( $rows = mysql_fetch_array($result))

{
  foreach ($rows as $vak)
{ 
   if ($val != "") { echo $val;}
}
}

edit :

The real fetched array is like this :

array(0 => data1,name_of_field1 => data1, 1 => data2,name_of_field2 => data2 , 3 => data3,name_of_field3 => data3)

you can access each value by a numeric key --> $rows[0] or by its field name --> $rows[name_of_field1] , both same.

notice the while loop, make it go for all the rows.

imarockstar · October 1, 2008

sorry ...

<?

$id = $_GET["id"];

//select the table
$result = mysql_query("select * from jobs where id='$id' ");

//grab all the content
while($r=mysql_fetch_array($result))
{	
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $jobtitle =$r["jobtitle"];
   $jobid =$r["jobid"];
   $recid =$r["recid"];
   $jobdesc =$r["jobdesc"];
   $jobreq1 =$r["jobreq1"];
   $jobreq2 =$r["jobreq2"];
   $jobreq3 =$r["jobreq3"];
   $jobreq4 =$r["jobreq4"];
   $jobreq5 =$r["jobreq5"];
   $jobreq6 = $r['jobreq6'];
   $jobreq7 = $r['jobreq7'];
   $jobreq8 = $r['jobreq8'];
   $jobreq9 = $r['jobreq9'];
   $jobreq10 = $r['jobreq10'];
   $jobreq11 = $r['jobreq11'];
   $jobreq12 = $r['jobreq12'];
   $jobloc =$r["jobloc"];
   $jobtype =$r["jobtype"];
   $jobpay =$r["jobpay"];
   $jobtravel =$r["jobtravel"];
?>
   

   
   




<h2 class=><?=$jobtitle ?></h2>

<h3 class=>Job Number: <?=$jobid ?></h3>



<h4 class=jobdisplay>Job Description</h4>

<span class=jobdescription>
<?php echo nl2br($jobdesc); ?>
</span>

<br><br>


<h4 class=jobdisplay>Carreer Requirments</h4>



<ul class=jobdata>
<li class=jobdata><?=$jobreq1 ?></li>
<li class=jobdata><?=$jobreq2 ?></li>
<li class=jobdata><?=$jobreq3 ?></li>
<li class=jobdata><?=$jobreq4 ?></li>
<li class=jobdata><?=$jobreq5 ?></li>
<li class=jobdata><?=$jobreq6 ?></li>
<li class=jobdata><?=$jobreq7 ?></li>
<li class=jobdata><?=$jobreq8 ?></li>
<li class=jobdata><?=$jobreq9 ?></li>
<li class=jobdata><?=$jobreq10 ?></li>
<li class=jobdata><?=$jobreq11 ?></li>
<li class=jobdata><?=$jobreq12 ?></li>
</ul>



<h4 class=jobdisplay>Job Location</h4>
<span class=jobdata><?=$jobloc ?></span>

<br><br>

<h4 class=jobdisplay>Employment Type</h4>
<span class=jobdata><?=$jobtype ?></span>

<br><br>

<h4 class=jobdisplay>Pay Rate</h4>
<span class=jobdata><?=$jobpay ?></span>

<br><br>

<h4 class=jobdisplay>Travel Required</h4>
<span class=jobdata><?=$jobtravel ?></span>

<br><br>



<div class=applynow>
<a href="jobs_apply?jobid=<?=$jobid ?>&jobtitle=<?=$jobtitle ?>&recid=<?=$recid ?>">
<img src="images/apply_now.gif" alt="image" width="273" height="118" />
</a>
</div>			



<?php } ?>

this is my main concern .. .these are always left blank ...

how would i not display them if there was no data ..

<ul class=jobdata>
<li class=jobdata><?=$jobreq1 ?></li>
<li class=jobdata><?=$jobreq2 ?></li>
<li class=jobdata><?=$jobreq3 ?></li>
<li class=jobdata><?=$jobreq4 ?></li>
<li class=jobdata><?=$jobreq5 ?></li>
<li class=jobdata><?=$jobreq6 ?></li>
<li class=jobdata><?=$jobreq7 ?></li>
<li class=jobdata><?=$jobreq8 ?></li>
<li class=jobdata><?=$jobreq9 ?></li>
<li class=jobdata><?=$jobreq10 ?></li>
<li class=jobdata><?=$jobreq11 ?></li>
<li class=jobdata><?=$jobreq12 ?></li>
</ul>

volatileboy · October 1, 2008

Try:

{
 foreach ($rows as $val)
{ 
  if (!empty($val)) { echo $val;}
}
}

F1Fan · October 1, 2008

Replace each <li> line with this:

<?php
if (!empty($jobreq1) echo "<li class=jobdata>$jobreq1</li>";
?>

volatileboy · October 1, 2008

Just a pointer, can't be bothered doing all of it lol, you could do this:

Instead of just putting this:

<li class=jobdata><?=$jobreq1 ?></li>

try this:

if(!empty($jobreq1)) {
   echo '<li class=jobdata>' . $jobreq1 . '</li>'
}

asmith · October 1, 2008

Put your while loop where you are echoing the list . AND you don't really need to put each $r[] variable in a new variable :

echo "<ul>";
while($r=mysql_fetch_array($result))
{

foreach($r as $val)
{
if ($val != "")
{
echo "<li class=jobdata>$jobreq1</li>";
}}}

echo "</ul>";

imarockstar · October 1, 2008

I did the above and it displays nothing ..

F1Fan · October 1, 2008

Which one? You've had a lot of options. Also, re-post your code now that you've changed it.

imarockstar · October 2, 2008

sorry i meant the one directly above .. here is my code ..

none of the job requirements are showing up .. blah

<?

$id = $_GET["id"];

//select the table
$result = mysql_query("select * from jobs where id='$id' ");

//grab all the content
while($r=mysql_fetch_array($result))
{	
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $jobtitle =$r["jobtitle"];
   $jobid =$r["jobid"];
   $recid =$r["recid"];
   $jobdesc =$r["jobdesc"];
   $jobreq1 =$r["jobreq1"];
   $jobreq2 =$r["jobreq2"];
   $jobreq3 =$r["jobreq3"];
   $jobreq4 =$r["jobreq4"];
   $jobreq5 =$r["jobreq5"];
   $jobreq6 = $r['jobreq6'];
   $jobreq7 = $r['jobreq7'];
   $jobreq8 = $r['jobreq8'];
   $jobreq9 = $r['jobreq9'];
   $jobreq10 = $r['jobreq10'];
   $jobreq11 = $r['jobreq11'];
   $jobreq12 = $r['jobreq12'];
   $jobloc =$r["jobloc"];
   $jobtype =$r["jobtype"];
   $jobpay =$r["jobpay"];
   $jobtravel =$r["jobtravel"];
?>
   

   
   




<h2 class=><?=$jobtitle ?></h2>

<h3 class=>Job Number: <?=$jobid ?></h3>



<h4 class=jobdisplay>Job Description</h4>

<span class=jobdescription>
<?php echo nl2br($jobdesc); ?>
</span>

<br><br>


<h4 class=jobdisplay>Carreer Requirments</h4>


<?php
echo "<ul>";
while($r=mysql_fetch_array($result))
{

foreach($r as $val)
{
if ($val != "")
{
echo "<li class=jobdata>$jobreq1</li>";
}}}

echo "</ul>";
?>


<h4 class=jobdisplay>Job Location</h4>
<span class=jobdata><?=$jobloc ?></span>

<br><br>

<h4 class=jobdisplay>Employment Type</h4>
<span class=jobdata><?=$jobtype ?></span>

<br><br>

<h4 class=jobdisplay>Pay Rate</h4>
<span class=jobdata><?=$jobpay ?></span>

<br><br>

<h4 class=jobdisplay>Travel Required</h4>
<span class=jobdata><?=$jobtravel ?></span>

<br><br>



<div class=applynow>
<a href="jobs_apply?jobid=<?=$jobid ?>&jobtitle=<?=$jobtitle ?>&recid=<?=$recid ?>">
<img src="images/apply_now.gif" alt="image" width="273" height="118" />
</a>
</div>			



<?php } ?>

F1Fan · October 2, 2008

Is you query supposed to return one row, or multiple rows?

asmith · October 3, 2008

remove this :

while($r=mysql_fetch_array($result))
{	
   //the format is $variable = $r["nameofmysqlcolumn"];
   //modify these to match your mysql table columns
  
   $jobtitle =$r["jobtitle"];
   $jobid =$r["jobid"];
   $recid =$r["recid"];
   $jobdesc =$r["jobdesc"];
   $jobreq1 =$r["jobreq1"];
   $jobreq2 =$r["jobreq2"];
   $jobreq3 =$r["jobreq3"];
   $jobreq4 =$r["jobreq4"];
   $jobreq5 =$r["jobreq5"];
   $jobreq6 = $r['jobreq6'];
   $jobreq7 = $r['jobreq7'];
   $jobreq8 = $r['jobreq8'];
   $jobreq9 = $r['jobreq9'];
   $jobreq10 = $r['jobreq10'];
   $jobreq11 = $r['jobreq11'];
   $jobreq12 = $r['jobreq12'];
   $jobloc =$r["jobloc"];
   $jobtype =$r["jobtype"];
   $jobpay =$r["jobpay"];
   $jobtravel =$r["jobtravel"];

and change

echo "<ul>";
while($r=mysql_fetch_array($result))
{

foreach($r as $val)
{
if ($val != "")
{
echo "<li class=jobdata>$jobreq1</li>";
}}}

echo "</ul>";

to

echo "<ul>";
while($r=mysql_fetch_array($result))
{

foreach($r as $val)
{
if ($val != "")
{
echo "<li class=jobdata>$val</li>";
}}}

echo "</ul>";

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