tlavelle Posted August 24, 2007 Share Posted August 24, 2007 There are 2 if statements at the beginning of this code snippet. The one that is currently commented out works while the one that is not commented out yields this error Parse error: syntax error, unexpected '{' in C:\xampp\htdocs\bench2\index4.php on line 106 I am just not seeing why. Help please? //if((isset($_POST['wercbench2_next']) and ($_POST['year']='2007')) or $_POST['year']='2005'){ if((isset($_POST['wercbench2_next']) and ($_POST['year']='2007')) or (isset($_POST['wercbench1_next']) and ($_POST['year']='2005')) { $useryear=$_POST['year']; $_SESSION['sess_year']=$_POST['year']; $userindustry=$_POST['industry']; $_SESSION['sess_ind']=$_POST['industry']; $metric_query = "SELECT metric_desc.met_desc, bench_data.median, bench_data.best_pract FROM metric_desc, bench_data where metric_desc.met_key=bench_data.met_key and $useryear=bench_data.year"; //$metric_query = "SELECT metric_desc.met_desc, bench_data.median, bench_data.best_pract FROM metric_desc, bench_data where metric_desc.met_key=bench_data.met_key and $useryear=bench_data.year and $userindustry=bench_data.comp_key"; $metric_result = mysql_query($metric_query) or die ("Error in query: $query. ".mysql_error()); if(isset($_POST['wercbench2_next']) and (mysql_num_rows($metric_result)) > 0){ // yes // print them one after another //echo "<form action=\"results.php\" method=\"post\" name\"wercbench2\">"; echo "<form action=\"results.php\" method=\"post\" name\"wercbench3\">"; echo "<table cellpadding=10 border=1>"; echo "<tr>"; echo "<td>Metric Description</td>"; echo "<td>Average</td>"; echo "<td>Best Practice</td>"; echo "<td>Enter your value here</td>"; echo "</tr>"; while($row = mysql_fetch_row($metric_result)) { echo "<tr>"; echo "<td>".$row[0]."</td>"; echo "<td>" . $row[1]."</td>"; echo "<td>".$row[2]."</td>"; echo "<td><input type=\"text\" name=\"myvalue[]\" value=\"\" /></td>"; echo "</tr>"; } echo "<tr><td colspan=\"3\"> </td><td align=\"center\"><input type=\"submit\" name=\"wercbench3\" value=\"Compare\"></td></tr></table>"; } // free result set memory mysql_free_result($year_result); //mysql_free_result($industry_result); mysql_free_result($metric_result); //mysql_free_result($metric_result); } Quote Link to comment Share on other sites More sharing options...
lemmin Posted August 24, 2007 Share Posted August 24, 2007 The problem is in the code above it, I believe. You also might want to change the '=' to '==' if you want to compare those two and not set them. Quote Link to comment Share on other sites More sharing options...
akitchin Posted August 24, 2007 Share Posted August 24, 2007 count the number of parentheses. you're missing a closing one: //if((isset($_POST['wercbench2_next']) and ($_POST['year']='2007')) or (isset($_POST['wercbench1_next']) and ($_POST['year']='2005')) { // 12 3 2 3 21 2 3 2 3 21 still left with one open parenthesis at the end. EDIT: lemmin is right, you'll need two = signs to compare the two. Quote Link to comment Share on other sites More sharing options...
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