learningcurve Posted June 20, 2012 Share Posted June 20, 2012 I have a query that is working beautifully (thanks to some stellar help form this forum). At this point, there is only one problem. I want a statement to print in the results right before a table that contains the results "Your query returned $num_rows responses." Right now it is printing the statement as such "Your query returned responses." and not even leaving a place where num_rows should be. I am not getting any errors either. Any ideas? I am fairly new to php, soooo..... <html> <head> </head> <body> <?php //db connections or error returns $connect = mysql_connect(DELETED); if (!$connect) { exit('<p>Unable to connect to the server at this time.</p>'); } mysql_select_db("surveys"); if (!mysql_select_db("surveys")) { exit ('<p>Unable to locate the database at this time.</p>'); } //if the "submit" variable does not exist, the form has not been submitted - display initial page if (!isset($_POST['submit'])) { //form ?> <form name="sgid" id="sgid" method="post" action="sgidsurveyquery.php"> <h1>SGID Satisfaction Survey Results</h1> <h2>Results can be filtered by evaluation party and semester/year given.</h2></br> Please choose: <?php //pull evaluator drop down choices from db as array $query="SELECT q1 FROM surveys.SGID_satisfaction"; $result=mysql_query($query); $options=""; while ($row=mysql_fetch_array($result)){ $evaluator=$row("q1"); $options.="<OPTION VALUE=>$evaluator</option"; } ?> <select name="evaluator"> <option value=0> <? echo $options?> </select> for <?php //pull date drop down choices from db as array $query="SELECT q2 FROM surveys.SGID_satisfaction"; $result=mysql_query($query); $options=""; while ($row=mysql_fetch_array($result)){ $date=$row("q2"); $options.="<OPTION VALUE=>$date</option"; } ?> <select name="date"> <option value=0> <? echo $options?> </select> <input type="submit" name ="submit" value="Show Results"> </form> <?php } else { //display results print"<h2>SGID Satisfaction Survey Results</h2><br/>"; $evaluator=$_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"; echo "<h3>Results of SGID Satisfaction Survey for $evaluator and $date</h3></br>"; $num_rows = mysql_num_rows($result); echo "Your query returned $num_rows responses."; //Start a table print "<table CELLPADDING=10 border =1 >"; print "<tr>"; print "<th style='width:150px'>Which organization did your SGID?</th>"; print "<th style='width:100px'>Please specify the semester and year the SGID was done.</th>"; print "<th style='width:300px'>What about this service was helpful for your teaching?</th>"; print "<th style='width:300px'>What about this service could be improved to support your teaching?</th>"; print "<th style='width:300px'>Would you be willing for us to contact you later about what, if anything, you have changed and how that has affected student learning?</th>"; print "</tr>"; $result = mysql_query ($query) or die(mysql_error()); while ($row = @mysql_fetch_array($result)) { print "<table CELLPADDING=10 border = 1>"; print "<tr>"; print "<td style='width:150px'>".$row['q1']."</td>"; print "<td style='width:100px'>".$row['q2']."</td>"; print "<td style='width:300px'>".$row['q3']."</td>"; print "<td style='width:300px'>".$row['q4']."</td>"; print "<td style='width:300px'>".$row['q5']."</td>"; print "</tr>"; print "</table>"; } //close results mysql_close(); } ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
nibbrwebdevelopment Posted June 20, 2012 Share Posted June 20, 2012 $query = "SELECT q1, q2, q3, q4, q5 FROM surveys.SGID_satisfaction WHERE q1='$evaluator' AND q2= '$date' ORDER BY created"; echo "<h3>Results of SGID Satisfaction Survey for $evaluator and $date</h3></br>"; $num_rows = mysql_num_rows($result); echo "Your query returned $num_rows responses."; By the looks of it here, your not getting the number of rows because $result variable does not exist; you have not ran the query! After $query = you need to run $result = mysql_query($query); Quote Link to comment Share on other sites More sharing options...
learningcurve Posted June 20, 2012 Author Share Posted June 20, 2012 Thanks for the clue! You were halfway right. I was running the query so that wasn't the problem. I was just asking for the variable $result before I had created/stated it. D'oh! Quote Link to comment Share on other sites More sharing options...
nibbrwebdevelopment Posted June 21, 2012 Share Posted June 21, 2012 lol Quote Link to comment Share on other sites More sharing options...
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