Re: trouble with MySQL 'LIKE' command

From: Date: Sat, 05 May 2001 07:00:26 +0000
Subject: Re: trouble with MySQL 'LIKE' command
References: 1 2 3 4  Groups: php.windows 
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Man you got me in the scripting mood... I think this is pretty useful to have on a site <? //searchDB.php mysql_connect($host,$user,$pw); echo genSearch($table,$ID,$search,$host,$user,$pw,$db); if(isset($search)){ echo genData($db,$table,$ID,$search); } /*********** fucntions ********/ function genSearch($table,$ID,$search,$host,$user,$pw,$db){ $dbs=mysql_list_dbs(); $num_db=mysql_num_rows($dbs); $ddb="<select name=db>"; for($d=0;$d<$num_db;$d++){ $ddb.="<option ".selected($db,mysql_db_name($dbs,$d)).">".mysql_db_name($dbs,$d)."</option> "; } $ddb.="</select>"; @mysql_select_db($db); $tbs=@mysql_list_tables($db); $num_tb=@mysql_num_rows($tbs); $tb="<select name=table>"; for($d=0;$d<$num_tb;$d++){ $tb.="<option ".selected($table,@mysql_tablename($tbs,$d)).">".@mysql_tablename($tbs,$d)." </option>"; } $tb.="</select>"; $fl=@mysql_list_fields($db,$table); $num_fd=@mysql_num_fields($fl); $fd="<select name=ID>"; for($d=0;$d<$num_fd;$d++){ $fd.="<option ".selected($ID,@mysql_field_name($fl,$d)).">".@mysql_field_name($fl,$d)."</o ption>"; } $fd.="</select>"; $r="<table bgcolor='#D0D0D0'>"; $r.="<tr><td><form method=post action='searchDB.php'></td></tr>"; $r.="<tr><td>Host</td><td><input type=text name=host value=$host></td></tr>"; $r.="<tr><td>User</td><td><input type=text name=user value=$user></td></tr>"; $r.="<tr><td>Password</td><td><input type=password name=pw value=$pw></td></tr>"; $r.="<tr><td>Database</td><td> $ddb </td></tr>"; $r.="<tr><td>Table</td><td> $tb </td></tr>"; $r.="<tr><td>Field</td><td> $fd </td></tr>"; $r.="<tr><td>Search</td><td><input type=text name=search value=$search></td></tr>"; $r.="<tr><td></td></tr><tr><td><input type=submit></form></td></tr></table>"; return $r; } function selected($a,$b){ return ($a==$b)?"selected":""; } function genData($db,$table,$ID,$search){ $fields =@mysql_list_fields($db,$table);//could have $link as 3rd argument $num=@mysql_num_fields($fields); $r ="<br><table><tr>"; for($f=0;$f<$num;$f++){ $r.="<td bgcolor='#D0DCE0'>".@mysql_field_name($fields,$f)."</td>"; } $r.="</tr>"; $sql=" select * from $table where lower($ID) like '%".strtolower($search)."%' "; $sql_result=@mysql_query($sql); $c=0; while($result=@mysql_fetch_array($sql_result)){ $c++; $r.="<tr> "; for($ii=0;$ii<count($result);$ii++){ $color=($c%2==0)?'#D0DCE0':'#DDDDDD'; $r.= " <td bgcolor=$color> ".htmlSpecialChars(stripslashes($result[$ii]))." </td> "; } $r.="</tr> "; } $r.="</table>"; return "<br><b>$sql</b><br><br> Matches: ".$c."<br>".$r; } ?> > ----- Original Message ----- > From: "r.gelstharp" <r.gelstharp@ntlworld.com> > To: <php-windows@lists.php.net> > Sent: Saturday, May 05, 2001 2:28 AM > Subject: Re: [PHP-WIN] trouble with MySQL 'LIKE' command > > > > Done like you recommended with the select....like '%".$var."%' but it > still > > gives me empty rows of data. > > > > Everytime I try and search for something it returns with 7 rows of data, > all > > of which are blank. I'm using the basic mysql_num_rows to get the number > of > > rows of data it retrieves, and I'm using a WHILE($data = > > mysql_fetch_array($sql_result)) to output the rows of data. > > > > If it's not the SQL statement, what could be doing this? I ask in case > > you've heard this problem mentioned by some thick newbie like meself > before. > > > > > > > > -- > > PHP Windows Mailing List (http://www.php.net/) > > To unsubscribe, e-mail: php-windows-unsubscribe@lists.php.net > > For additional commands, e-mail: php-windows-help@lists.php.net > > To contact the list administrators, e-mail: php-list-admin@lists.php.net > > > > > > -- > PHP Windows Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-windows-unsubscribe@lists.php.net > For additional commands, e-mail: php-windows-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net > >

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