kaimason1 Posted March 1, 2009 Share Posted March 1, 2009 Unknown column 'Book' in 'where clause' Now I'm getting this error... help? EDIT:: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML> <HEAD> <TITLE>My Search</TITLE> <META NAME="Generator" CONTENT="EditPlus"> <META NAME="Author" CONTENT=""> <META NAME="Keywords" CONTENT=""> <META NAME="Description" CONTENT=""> </HEAD> <BODY> <FORM METHOD="get" action="search.php"> <P>Please enter your search words here and specify the type of data you input: <INPUT type="text" name="search" size=30 maxlength=70> <SELECT name="type" size="1"> <OPTION value="Title">Title</OPTION> <OPTION value="Author">Author</OPTION> <OPTION value="Genre">Genre</OPTION> <OPTION value="Page_Count">Page Count</OPTION> <OPTION value="Reading_Level">Reading Level</OPTION> <OPTION value="ISBN">ISBN</OPTION> </SELECT> </P> <INPUT type="submit" name="submit" value="Submit"> </FORM> </BODY> </HTML> <?php if(isset($_GET['submit']) && $_GET['submit'] == "Submit") { //Connect to database// include("usernamepassworddatabase.php"); mysql_connect(localhost,$username,$password)or die("I couldnt connect, sir. By the way, am I being paid for acting like a personal butler?"); mysql_select_db($database)or die("The database you are trying to access is currently unavailable. Curious..."); //Retrieve variables from HTML form $search=trim(mysql_real_escape_string($_GET['search'])); $type=trim(mysql_real_escape_string($_GET['type'])); //Create MySQL query// if($search == "") { die("Please enter a search."); } else if($type == "Title") { $query="SELECT * FROM TEST1_contacts WHERE 'Book' LIKE '%$search%'"; } else if($type == "Author") { $query="SELECT * FROM TEST1_contacts WHERE 'Author' LIKE '%$search%'"; } else if($type == "Genre") { $query="SELECT * FROM TEST1_contacts WHERE 'Genre' LIKE '%$search%'"; } else if($type == "Page_Count") { $query="SELECT * FROM TEST1_contacts WHERE 'Page_Count' LIKE '%$search%'"; } else if($type == "Reading_Level") { $query="SELECT * FROM TEST1_contacts WHERE 'Reading_Level' LIKE '%$search%'"; } else if($type == "ISBN") { $query="SELECT * FROM TEST1_contacts WHERE 'ISBN' LIKE '%$search%'"; } else{ die("Congratulations -- you achieved the impossible rating of hacker. How else did I receive an inaccurate data type, hmm?"); } //Search Results// $result=mysql_query($query) or die (mysql_error()); $num=mysql_num_rows($result); if($num == 0) { die("There seems to be no records matching your search. Interesting..."); } else{ $i=0; while($i<$num){ $book=mysql_result($result,$i,"Book"); $author=mysql_result($result,$i,"Author"); $genre=mysql_result($result,$i,"Genre"); $page_count=mysql_result($result,$i,"Page_Count"); $reading_level=mysql_result($result,$i,"Reading_Level"); $ISBN=mysql_result($result,$i,"ISBN"); echo("<b>$book</b> by $author.<br>$genre.<br>$page_count pages.<br>$reading_level reading level.<br>ISBN:$ISBN<br><hr><br>"); ++$i; } } } php?> Quote Link to comment Share on other sites More sharing options...
Philip Posted March 1, 2009 Share Posted March 1, 2009 The column 'Book' does not exist in your table, TEST1_contacts, on this line: $query="SELECT * FROM TEST1_contacts WHERE 'Book' LIKE '%$search%'"; Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted March 1, 2009 Share Posted March 1, 2009 Putting single-quotes around column names make them strings instead of identifiers. Remove the single-quotes from around all your column names. Quote Link to comment Share on other sites More sharing options...
RussellReal Posted March 1, 2009 Share Posted March 1, 2009 pfm is right, use ` ` instead of ' ' Quote Link to comment Share on other sites More sharing options...
kaimason1 Posted March 1, 2009 Author Share Posted March 1, 2009 I just used the wrong table name... oops Quote Link to comment Share on other sites More sharing options...
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