php/mysql query not showing results in a table

learningcurve · June 14, 2012

I have a form(sgid5form.php):

<form name="sgid" id="sgid" method="post" action="sgid5.php">
<h1>Results from SGID satisfaction survey.  Results can be filtered by evaluation party or semester/year given.</h1></br>
Please choose:
<select name="evaluator">
<option value="ALT		(John, Tim and Bob)">ALT		(John, Tim and Bob)</option>
<option value="CELTUA 		(Geoff or Joe)">CELTUA 		(Geoff or Joe)</option>
<option value="CTL MUM		(MUM Faculty)">CTL MUM		(MUM Faculty)</option>
<option value="CTL VOA		(VOA or MUM Faculty)">CTL VOA		(VOA or MUM Faculty)</option>
</select>

for
<select name="date">
<option value="Fall 2010">Fall 2010</option>
<option value="Fall 2011">Fall 2011</option>
<option value="Fall 2012">Fall 2012</option>
<option value="Fall 2013">Fall 2013</option>
<option value="Spring 2010">Spring 2010</option>
<option value="Spring 2011">Spring 2011</option>
<option value="Spring 2012">Spring 2012</option>
<option value="Spring 2012">Spring 2013</option>
</select>
<input type="submit" name ="submit" value="Show Results">
</form>

That submits to this php page (sgid5.php):

<html>
<head>
</head>
<body>
<?php
$connect = mysql_connect(#####DELETED####);
mysql_select_db("surveys");
print "SGID Satisfaction Survey Results";

$evaluator=$_POST['evaluator'];
$date=$_POST['date'];

$query = "SELECT q1, q2, q3, q4, q5
FROM surveys.SGID_satisfaction
WHERE q1='$evaluator'
AND q2= '$date'
ORDER BY created";

$result = mysql_query ($query);
while ($row = @mysql_fetch_array($result))

//Start a table
print "<center>";
print "<table CELLPADDING=10 border =1 >";
print "<tr>";
print "<th>q1</th>";
print "<th>q2</th>";
print "<th>q3</th>";
print "<th>q4</th>";
print "<th>q5</th>";
print "</tr>";
print "<tr>";
print "<td>".$row['q1']."</td>";
print "<td>".$row['q2']."</td>";
print "<td>".$row['q3']."</td>";
print "<td>".$row['q4']."</td>";
print "<td>".$row['q5']."</td>";
print "</tr>";
print "</table>";

mysql_close();
?>
</body>
</html>

That should return the results in a real simple table. I am not getting any errors, but no results are posting either. I have attached a jpeg of what appears on my screen. I am really new to php and need all the help I can get.

Jessica · June 14, 2012

You're suppressing errors with the @ symbol. Remove that, and ensure error reporting is on and set to E_ALL.

learningcurve · June 14, 2012

Okay, thanks for the heads up. I removed the @. Errors were already turned on and I was getting them earlier in my coding. After removing the @, condition is the - no errors and same output.

thanks!

scootstah · June 14, 2012

Did you make sure it is even returning anything? You can check with mysql_num_rows($result);.

mrMarcus · June 14, 2012

Try echo'ing your $query to make sure it's what you're expecting:

$query = "SELECT q1, q2, q3, q4, q5
FROM surveys.SGID_satisfaction
WHERE q1='$evaluator'
AND q2= '$date'
ORDER BY created";

echo $query;

And check for query errors:

$result = mysql_query ($query) or die(mysql_error());

Jessica · June 14, 2012

echo your $query before running it and see if it looks the way you expect. Try running it in a tool like phpmyadmin and see if it works. What does your data look like

learningcurve · June 14, 2012

Great suggestions!

echo returns just what it should and looks good

running the query in sequel pro returns results when I substitute the variable for one of the choices

adding mysql_num_rows tells me the query created six rows for the example I ran where I knew data would return.

Any more suggestions? :'(

scootstah · June 14, 2012

Can you eliminate other output in your script and post back the exact results of this?

$query = "SELECT q1, q2, q3, q4, q5
FROM surveys.SGID_satisfaction
WHERE q1='$evaluator'
AND q2= '$date'
ORDER BY created";

$result = mysql_query ($query) or die('Query: ' . $query . '<br>Error: ' . mysql_error());

echo 'Query: ' . $query . '<br>';
echo 'Rows returned: ' . mysql_num_rows($result) . '<br>';

echo '<pre>';
while($row = mysql_fetch_assoc($result)) {
print_r($row);
}
echo '</pre>';

mrMarcus · June 14, 2012

Great suggestions!

echo returns just what it should and looks good

running the query in sequel pro returns results when I substitute the variable for one of the choices

adding mysql_num_rows tells me the query created six rows for the example I ran where I knew data would return.

Any more suggestions? :'(

You shouldn't need to substitute anything. Take exactly what was echo'd to the screen and put it directly in your sequel pro.

learningcurve · June 14, 2012

Scootstah,

Did what you wanted and here is what came back. Looks good - just what I want to appear in my table.

SGID Satisfaction Survey ResultsQuery: SELECT q1, q2, q3, q4, q5 FROM surveys.SGID_satisfaction WHERE q1='CELTUA (Geoff or Joe)' AND q2= 'Spring 2012' ORDER BY created
Rows returned: 4

Array
(
    [q1] => CELTUA 		(Geoff or Joe)
    [q2] => Spring 2012
    [q3] => This service is very useful; it is much more useful than the end-semester evaluations. In particular, the strength and weakness of the teaching are given in very detail, thus it is very easy to follow the suggestions to make improvement in teaching. I have absolutely benefited from this program.  
    [q4] => It is almost perfect.
    [q5] => yes, I would like to be contacted. ;;
)
Array
(
    [q1] => CELTUA 		(Geoff or Joe)
    [q2] => Spring 2012
    [q3] => Discussing the 'improvement' part of the comments with SGID facilitator was helpful.  The facilitator gave me some suggestions for improvements based on his experience.
    [q4] => 
    [q5] => Yes. ;;
)
Array
(
    [q1] => CELTUA 		(Geoff or Joe)
    [q2] => Spring 2012
    [q3] => Getting a mid-term update on what the class is thinking.
    [q4] => 
    [q5] => 
)
Array
(
    [q1] => CELTUA 		(Geoff or Joe)
    [q2] => Spring 2012
    [q3] => Helpful feedback. It was also helpful to discuss the results in-person with the SGID leader.
    [q4] => The only thing that I can think of is maybe to advertise it more widely so that more people know about it. Most of my students have commented that they have never done this before. ;;  ;; Also, it may be helpful to have it take less than 1/2 hr. or have choices for a 1/2 hr session or a 15 min. session. ;;  ;; I would also be interested to know what students think about it.
    [q5] => 
)

So what does this mean? I am really, really appreciating all the help I am getting here.

scootstah · June 14, 2012

Upon closer inspection it appears you did not use any opening/closing brackets on your while loop, which I believe is your problem. Does this work?

while ($row = @mysql_fetch_array($result)) {
//Start a table
print "<center>";
print "<table CELLPADDING=10 border =1 >";
print "<tr>";
print "<th>q1</th>";
print "<th>q2</th>";
print "<th>q3</th>";
print "<th>q4</th>";
print "<th>q5</th>";
print "</tr>";
print "<tr>";
print "<td>".$row['q1']."</td>";
print "<td>".$row['q2']."</td>";
print "<td>".$row['q3']."</td>";
print "<td>".$row['q4']."</td>";
print "<td>".$row['q5']."</td>";
print "</tr>";
print "</table>";
}

learningcurve · June 14, 2012

You, sir, are an absolute fricking genius! That was it. The table is kinda wobbly and ugly looking, but I can tidy it up now that it is working. BRAVO and THANK YOU!

scootstah · June 14, 2012

All that work for missing brackets...

Oh, and as a side note: you don't need to (and shouldn't) run mysql_close(). PHP will automatically close the MySQL connection for you.

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php/mysql query not showing results in a table

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