angelsRock Posted September 4, 2007 Share Posted September 4, 2007 i get the value from drop down box.. when i display echo it .. i can view all the data i selected.... but when i get these data apply into the query below .. it did not show the data i want.. <?php if(isset($_POST[submitform])){ $location= $_POST[location]; $category= $_POST[category]; $option= $_POST[option]; echo"$location"; echo"$category"; echo"$option"; }?> <?php $TableName = "prem_propertyad"; $Conn =mysql_connect($server, $user, $pass); mysql_select_db($db); $Query = "SELECT * from $TableName where type1='$option'"; if ($location !="All Locations") $Query = $Query. " and state= '$Location'"; if ($category !="All Category") $Query = $Query. " and type2='$category'"; $Result=mysql_query($Query); print ("<TABLE BORDER=1 WIDTH=\"750\" CELLSPACING=2 CELLPADDING=2 ALIGN=center>\n"); print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>Sale/Rent</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Location</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Price(RM)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Type</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Build Up Area/Land Area (sq ft)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Date Posted</TD>\n"); print ("</TR>\n"); while ($Row = mysql_fetch_array ($Result)) { print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[type1] </TD>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[state] </TD>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[selling_price]</TD>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[type3]</TD>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[buildup_area]</TD>\n"); print ("<TD ALIGN=center VALIGN=top>$Row[datePosted]</TD>\n"); print ("</TR>\n"); } mysql_close($Conn); print ("</TABLE>\n"); ?> Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/ Share on other sites More sharing options...
darkfreaks Posted September 4, 2007 Share Posted September 4, 2007 Take out the Quotes and change the Query <?php if(isset($_POST[submitform])){ $location= $_POST[location]; $category= $_POST[category]; $option= $_POST[option]; echo ($location); echo($category); echo ($option); } $Query = ("SELECT * FROM ".$TableName." WHERE type1='$option'"); ?> Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341510 Share on other sites More sharing options...
lemmin Posted September 4, 2007 Share Posted September 4, 2007 You are getting a syntax error? What is this error? You really should use literal strings when using a key in an array, like $_POST['location']. Try $Result=mysql_query($Query) or die(mysql_error()); To see the error. If there is no error, there probably isn't an entry that has $option for "type1" in table "prem_propertyad". Check that the entry that you want exists. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341518 Share on other sites More sharing options...
darkfreaks Posted September 4, 2007 Share Posted September 4, 2007 Corrected Query: <?php $Query = ("SELECT * FROM ".$TableName." WHERE type1=".$option.""); ?> Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341521 Share on other sites More sharing options...
lemmin Posted September 4, 2007 Share Posted September 4, 2007 Variables are parsed in double quotes. I don't think your query is any different than the one he posted. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341522 Share on other sites More sharing options...
darkfreaks Posted September 4, 2007 Share Posted September 4, 2007 i escaped the variables so they could be read into the SQL otherwise it wont recognize the $table variable Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341524 Share on other sites More sharing options...
lemmin Posted September 4, 2007 Share Posted September 4, 2007 PHP parses the variables (in double quotes) before it is sent to sql as a string. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341526 Share on other sites More sharing options...
darkfreaks Posted September 4, 2007 Share Posted September 4, 2007 either way i cleaned up her syntax abit Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341531 Share on other sites More sharing options...
darkfreaks Posted September 4, 2007 Share Posted September 4, 2007 Final Code: <?php $TableName = "prem_propertyad"; $Conn =mysql_connect($server, $user, $pass); mysql_select_db($db); $Query = ("SELECT * FROM ".$TableName." WHERE type1=".$option."" if ($location !="All Locations") $Query = $Query. " and state= ".$Location.""; if ($category !="All Category") $Query = $Query. " and type2=".$category.""; $Result=mysql_query($Query)or die(mysql_error("Could not select from database!")); print ("<TABLE BORDER=1 WIDTH=\"750\" CELLSPACING=2 CELLPADDING=2 ALIGN=center>\n"); print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>Sale/Rent</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Location</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Price(RM)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Type</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Build Up Area/Land Area (sq ft)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Date Posted</TD>\n"); print ("</TR>\n"); while ($Row = mysql_fetch_array ($Result)) { print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type1']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['state']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['selling_price']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type3']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['buildup_area']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['datePosted']."</TD>\n"); print ("</TR>\n"); } mysql_close($Conn); print ("</TABLE>\n"); ?> Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-341546 Share on other sites More sharing options...
angelsRock Posted September 5, 2007 Author Share Posted September 5, 2007 bro.. i had followed your way.. but now the whole page became blank... it din show any error msg.. just blank whole page... i had unchekced the show friendly messaage ..below is my full code.. i think of no idea why show wrong din data that i wan... [quote]<?php include('./uservar.php'); ?> <?php error_reporting(E_ALL); if(isset($_POST[submitform])){ $location= $_POST[location]; $category= $_POST[category]; $option= $_POST[option]; echo"$location"; echo"$category"; echo"$option"; }?> <html> <head> <meta http-equiv="Content-Language" content="en-us"> <meta http-equiv="Content-Type" content="text/html; charset=windows-1252"> <title>Neuproperty</title> <style fprolloverstyle>A:hover {color: #FFFFCC} </style> <script type="text/javascript" language="JavaScript1.2" src="menu/stmenu.js"></script> </head> <body link="#FFFFFF" vlink="#FFFFFF" alink="#FFFFFF" bgcolor="#C0C0C0"> <div align="center"> <table border="0" style="border-collapse: collapse" width="950" id="table1"> <tr> <td width="566" rowspan="2" bgcolor="#FFFFFF"> <img border="0" src="img/logo.gif" width="235" height="67"></td> <td height="22" colspan="3" valign="top" bgcolor="#FFFFFF"> </td> </tr> <tr> <td colspan="3" height="47" bgcolor="#FFFFFF"> <p align="right"><b><font face="Arial" color="#333333" size="2"> Quick Property Search</font></b> <input type="text" name="T1" size="25"><input type="submit" value="Find" name="B1"></td> </tr> <tr> <td colspan="4" bgcolor="#FFFFFF"> <div align="center"> <table border="0" width="100%" cellspacing="0" cellpadding="0" id="table2"> <td height="28" colspan="0"><script type="text/javascript" language="JavaScript1.2" src="menu/property.js"></script> </td> <tr> <td bgcolor="#009696" height="8" width="6%"> </td> </tr> </table> <p align="left"><b><font face="Arial">Search</font></b></div> </td> </tr> <tr> <td colspan="3" bgcolor="#FFFCF9"> <FORM name="frmPropWanted" action="results.php" method=post> <TABLE cellSpacing=0 cellPadding=0 align="center" width="101%" bgColor=#dddddd border=0 id="table4" height="15"> <TR> `<TD height="15" bgcolor="#FFFCF9"> <select size="1" name="location"> <option selected>All Locations</option> <option>Kedah</option> <option>Perlis</option> <option>Pulau Pinang</option> <option>Perak</option> <option>Selangor</option> <option>Sabah</option> <option>Sabah</option> <option>Negeri Sembilan</option> <option>Melaka</option> <option>Johor</option> <option>Terrenganu</option> <option>Kelantan</option> <option>W.P. Kuala Lumpur</option> <option>W.P Labuan</option> </select> <select size="1" name="category"> <option selected>All Category</option> <option>Residential Property</option> <option>Commercial Property</option> <option>Industrial Property</option> <option>Land</option> <option>Room</option> </select> <select size="1" name="option"> <option>Rental</option> <option>Sales</option> <option selected>Rental / Sales</option> </select> <input type="submit" value="Search" name="submitform"> <input type="reset" value="Reset" name="B2"><br> </FORM> </TD> </TR> <TR> <TD height="15" bgcolor="#FFFFFF" colspan="3"> <table border="0" style="border-collapse: collapse" width="100%" cellpadding="0" id="table6"> <tr> <td bgcolor="#FFE2D9"> <font face="Arial" size="2"> records </font> </td> </tr> <tr></tr> </table> <div align="left"> </div> </tr> </TABLE> <?php error_reporting(E_ALL); $TableName = "prem_propertyad"; $Conn =mysql_connect($server, $user, $pass); mysql_select_db($db); $Query = ("SELECT * from ".$TableName." where type1=".$option.""; ) if ($location !="All Locations") $Query = $Query. " and state= ".$Location.""; if ($category !="All Category") $Query = $Query. " and type2=".$category.""; $Result=mysql_query($Query)or die(mysql_error("Could not select from database!")); print ("<TABLE BORDER=1 WIDTH=\"750\" CELLSPACING=2 CELLPADDING=2 ALIGN=center>\n"); print ("<TR ALIGN=center bgcolor=999999 VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>Sale/Rent</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Location</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Price(RM)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Type</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Build Up Area/Land Area (sq ft)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Date Posted</TD>\n"); print ("</TR>\n"); while ($Row = mysql_fetch_array ($Result)) { print ("<TR ALIGN=center bgcolor='pink' VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type1']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['state']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['selling_price']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type3']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['buildup_area']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['datePosted']."</TD>\n"); print ("</TR>\n"); } mysql_close($Conn); print ("</TABLE>\n"); ?> <br> <?php $TableName2 = "free_propertyad"; $Conn =mysql_connect($server, $user, $pass); mysql_select_db($db); $Query = ("SELECT * from ".$TableName2." where type1=".$option.""; ) if ($location !="All Locations") $Query = $Query. " and state= ".$Location.""; if ($category !="All Category") $Query = $Query. " and type2=".$category.""; $Result=mysql_query($Query)or die(mysql_error("Could not select from database!")); print ("<TABLE BORDER=1 WIDTH=\"750\" CELLSPACING=2 CELLPADDING=2 ALIGN=center>\n"); print ("<TR ALIGN=center bgcolor=999999 VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>Sale/Rent</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Location</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Price(RM)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Type</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Build Up Area/Land Area (sq ft)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Date Posted</TD>\n"); print ("</TR>\n"); while ($Row = mysql_fetch_array ($Result)) { print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type1']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['state']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['selling_price']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type3']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['buildup_area']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['datePosted']."</TD>\n"); print ("</TR>\n"); } mysql_close($Conn); print ("</TABLE>\n"); ?> <br> <?php $TableName3 = "basic_propertyad"; $Conn =mysql_connect($server, $user, $pass); mysql_select_db($db); $Query = ("SELECT * from ".$TableName3." where type1=".$option.""; ) if ($location !="All Locations") $Query = $Query. " and state= ".$Location.""; if ($category !="All Category") $Query = $Query. " and type2=".$category.""; $Result=mysql_query($Query)or die(mysql_error("Could not select from database!")); print ("<TABLE BORDER=1 WIDTH=\"750\" CELLSPACING=2 CELLPADDING=2 ALIGN=center>\n"); print ("<TR ALIGN=center bgcolor=999999 VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>Sale/Rent</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Location</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Price(RM)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Type</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Build Up Area/Land Area (sq ft)</TD>\n"); print ("<TD ALIGN=center VALIGN=top>Date Posted</TD>\n"); print ("</TR>\n"); while ($Row = mysql_fetch_array ($Result)) { print ("<TR ALIGN=center VALIGN=top>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type1']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['state']." </TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['selling_price']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['type3']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['buildup_area']."</TD>\n"); print ("<TD ALIGN=center VALIGN=top>".$Row['datePosted']."</TD>\n"); print ("</TR>\n"); } mysql_close($Conn); print ("</TABLE>\n"); ?> </td> </tr> <tr> <td colspan="2" bgcolor="#FFFCF9"> </td> <td width="379" bgcolor="#FFFCF9"> </td> </tr> <tr> <td colspan="4" bgcolor="#FFFFFF"> <table border="0" style="border-collapse: collapse" width="950" cellspacing="1" id="table3"> <tr> <td width="100%" height="20"><hr noshade size="1"></td> </tr> <tr> <td width="100%" height="26"> <p align="center"><b><font size="1" color="#808080">Help | Contact | About this site | Terms and Conditions | </font></b></td> </tr> </table> </td> </tr> </table> </body> </html>[/quote] Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342116 Share on other sites More sharing options...
angelsRock Posted September 6, 2007 Author Share Posted September 6, 2007 expert!!!.. help needed.. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342571 Share on other sites More sharing options...
darkfreaks Posted September 6, 2007 Share Posted September 6, 2007 change the code back to what you had before see if it works better Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342578 Share on other sites More sharing options...
trq Posted September 6, 2007 Share Posted September 6, 2007 You never check your query succeeds before attempting to use it. The general syntax for a select query should be something like... <?php $sql = "SELECT foo FROM bar"; if ($result = mysql_query($sql)) { if (mysql_num_rows($result)) { // it is now safe to display your retrieved data. while ($row = mysql_fetch_assoc($result)) { echo $row['foo']."<br />"; } } else { echo "No results found"; } } else { echo "Query failed<br />$sql<br />". mysql_error(); } ?> Apply this example to your code and let use know what the results are. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342580 Share on other sites More sharing options...
angelsRock Posted September 6, 2007 Author Share Posted September 6, 2007 my query able to display the data.. hi.. darkfreaks.. for the code u written.. [code=php:0] $Query = ("SELECT * FROM ".$TableName." WHERE type1=".$option."" if ($location !="All Locations") $Query = $Query. " and state= ".$Location.""; if ($category !="All Category") $Query = $Query. " and type2=".$category.""; my browser wrote an error at the line $Query = ("SELECT * FROM ".$TableName." WHERE type1=".$option."" error on ";" is it should be like this ?? $Query = ("SELECT * FROM ".$TableName." WHERE type1=".$option."") ; Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342626 Share on other sites More sharing options...
trq Posted September 6, 2007 Share Posted September 6, 2007 darkfreak's query is no good, use... $Query = "SELECT * FROM $TableName WHERE type1 = '$option'"; Did you change your code to reflect the example I posted? Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342641 Share on other sites More sharing options...
darkfreaks Posted September 6, 2007 Share Posted September 6, 2007 yeah i tested it my syntax wont work it just comes up blank you should follow thorpe's advice that is proper SQL structure. Quote Link to comment https://forums.phpfreaks.com/topic/67950-syntax-error/#findComment-342714 Share on other sites More sharing options...
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