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> Link to comment https://forums.phpfreaks.com/topic/264511-num_rows-not-showing/ 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); Link to comment https://forums.phpfreaks.com/topic/264511-num_rows-not-showing/#findComment-1355568 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! Link to comment https://forums.phpfreaks.com/topic/264511-num_rows-not-showing/#findComment-1355579 Share on other sites More sharing options...
nibbrwebdevelopment Posted June 21, 2012 Share Posted June 21, 2012 lol Link to comment https://forums.phpfreaks.com/topic/264511-num_rows-not-showing/#findComment-1355689 Share on other sites More sharing options...
Recommended Posts
Archived
This topic is now archived and is closed to further replies.