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PHP - Mysql Search Database With Multiple Columns
I have a search set up to search a table for the text entered in a textbox, I have two columns in the table, one with the first name of people, and the second with their last names, I am wondering how I can search both, so for instance:
I type in the search field: Roger Smith in the database it would look like: First_name-----|-----Last_name -------------------|------------------- Roger------------|-------Smith my current query is: Code: [Select] $query = mysql_query("SELECT * FROM users WHERE fname LIKE '%$find%' OR lname LIKE '%$find%'"); But if I type both parts of the name it doesn't return anything. works fine if I just search for "Roger" OR "Smith". Similar TutorialsIm having problems with this code in my search button script. I get the following error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource Here's the code I use on my search form. Code: [Select] <form method="post" action="srch_all.php"><input type=text name='search' size=15 maxlength=255><br><input type=submit></form> and here's the code I'm using to perform the search query: Code: [Select] if ($search) // perform search only if a string was entered. { mysql_connect($host, $user, $pass) or die ("Problem connecting to Database"); $srch="%".mysql_real_escape_string($search)."%"; $query = "select * from database WHERE column1 LIKE '$srch' or column2 LIKE '$srch' or column3 LIKE '$srch' or column4 LIKE '$srch' ORDER BY column1, column2 DESC, column3 ASC"; $result = mysql_db_query("database_name", $query); if(mysql_num_rows($result)==0) { print "<h2>Your search returned 0 Results</h2>"; } else if ($result) { <restult data stuff here>.. Cheers Hi all How do I modify the below code to search multiple tables in mySQL database? $query = "select * from store_items where description like \"%$trimmed%\" or title like \"%$trimmed%\" or dimensions like \"%$trimmed%\" order by id ASC"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); It is for a search function and I need it to search another table called 'about_text' and 'faq_text' Thanks Pete I made a small editing system for my news page, and I need to update three columns within my table "announcements" in the database. I tried a method of updating all of them with one MySQL query instead of using three as it just isn't neat. I've searched several methods via google and I've tried all of them, but just can't seem to get it to work. Is this MySQL query correct? mysql_query("UPDATE announcements SET title = {$title} WHERE id = '$id', content = {$content} WHERE id = '$id', lastmodified = ". date('M-d-Y') ." WHERE id = '$id'"); I am loading notifications from a database table called "notifications" and I am having a little trouble getting them to order in the correct way. my query I'm using right now: Code: [Select] $query = mysql_query("SELECT B.* FROM (SELECT A.* FROM notifications A WHERE A.user_id='$session' AND A.from_id!='$session' ORDER BY A.id ASC ) AS B ORDER BY B.state ASC LIMIT 7"); this works well as far as showing the unread notifications on top, then the read notifications below, however it's not ordering the two sets by ID (which the id auto increments so the higher id number is the newest) from newest on top to the oldest on bottom, still keeping them separated by the unread on top, read on bottom (column name is state for showing whether they're read or not). The order the notifications are displaying by their ID is: 3 5 2 4 1 when it should be: 5 3 2 4 1 Need some help I have 2 tables in a database and I need to search the first table and use the results from that search, to search another table, can this be done? and if it can how would you recommend that I go about it? Thanks For Your Help Guys! Hi, This has been baffling me for a couple hours now and i cant seem to figure it out. I have some code which creates an array and gets info from a mysql database and then displays in a list. This works great but after adding more and more rows to my database the list is now becoming quite large and doesnt look great on my site. Is it possible to split the list into multiple columns of about 25 and if possible once 3 or 4 columns have been created start another column underneath. To help explain i would be looking at a layout as follows: Code: [Select] line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25 line 1 line 1 line 1 line 2 line 2 line 2 ... ... ... line 25 line 25 line 25Im guessing there should be some sort of if statement to check how many items are being displayed and to create a new column if necessary. Is this correct? Thanks, Alex This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=313325.0 <?php if(isset($_POST['ActivityId'])) { require_once('database.php'); $Search=$_POST['Search']; $query="select * from registration where Flag='A' AND Activity like '%,$Search,%'"; $result=mysqli_query($dbc,$query) or die('Not Connected'); while($row=mysqli_fetch_array($result)) { echo "<table border='2'><tr><td> Name--".$row['Name']."</td> <td>Contact Number--".$row['ContactNumber']."</td> <td>Email--".$row['Email']."</td> <td>Address--".$row['Address']."</td></tr></table>"; } } elseif(isset($_POST['EventId'])){ require_once('database.php'); $Search=$_POST['Search']; $query="select * from participation where EventId ='$Search'"; $result=mysqli_query($dbc,$query) or die('Not Connected'); while($row=mysqli_fetch_array($result)){ $LoginId= $row['LoginId']; echo $LoginId;} $query="select * from registration where LoginId='$LoginId'"; $result=mysqli_query($dbc,$query) or die('Not Connected to Reg'); while($row=mysqli_fetch_array($result)){ echo "<table border='2'><tr><td> Name--".$row['Name']."</td> <td>Contact Number--".$row['ContactNumber']."</td> <td>Email--".$row['Email']."</td> <td>Address--".$row['Address']."</td></tr></table>"; }} elseif(isset($_POST['Location'])) { require_once('database.php'); $Search=$_POST['Search']; $query="select * from registration where Flag='A' AND Address like '%$Search%'"; $result=mysqli_query($dbc,$query) or die('Not Connected'); while($row=mysqli_fetch_array($result)) { echo "<table border='2'><tr><td> Name--".$row['Name']."</td> <td>Contact Number--".$row['ContactNumber']."</td> <td>Email--".$row['Email']."</td> <td>Address--".$row['Address']."</td></tr></table>"; } } elseif(isset($_POST['BloodGroup'])) { require_once('database.php'); $Search=$_POST['Search']; $query="select * from registration where Flag='A' AND BloodGroup ='$Search'"; $result=mysqli_query($dbc,$query) or die('Not Connected'); while($row=mysqli_fetch_array($result)) { echo "<table border='2'><tr><td> Name--".$row['Name']."</td> <td>Contact Number--".$row['ContactNumber']."</td> <td>Email--".$row['Email']."</td> <td>Address--".$row['Address']."</td></tr></table>"; } } ?> <html><body><form action='<?php echo $_SERVER['PHP_SELF']?>' method="post"> <input type="text" name="Search"><br> <input type="submit" name="ActivityId" value="Submit Activity Id"><br> <input type="submit" name="EventId" value="Submit Event Id"><br> <input type="submit" name="Location" value="Submit Location"><br> <input type="submit" name="BloodGroup" value="Submit Blood Group"><br> </form> </body></html> Hi frds............. I have two tables named "registration" and "participation". Registration table have some columns like "LoginId", "email","contact no.", "Name" etc.and participation table have 2 column named "LoginId" and "EventId". Now if user enters EventId, I want to display the details like email,contact no. etc. I have a code but it not works properly....... If result have more than one row it only displays the email and contact no. of last one instead of all. Plzzzzzzzzz help anyone................... ????????/ ugh... I'm a total PHP nub and I'm having trouble with: Code: [Select] $search_by = $_POST['search_by']; $search = $_post['search']; $dbc = mysqli_connect('xx', 'artofwarsomerset', 'xx', 'artofwarsomerset') or die ('Error connecting to MySQL server'); $query = "SELECT * FROM players WHERE '$search_by' = '$search' "; $result = mysqli_query($dbc,$query) or die("Error: ".mysqli_error($dbc)); echo "<table><tr><td>Player</td><td>city</td><td>alliance</td><td>x</td><td>y</td><td>other</td><td>porters</td><td>conscripts</td><td>Spies</td><td>HBD</td><td>Minos</td><td>LBM</td><td>SSD</td><td>BD</td><td>AT</td><td>Giants</td><td>Mirrors</td><td>Fangs</td><td>ogres</td><td>banshee</td></tr>" ; while ($row = mysqli_fetch_array ($result)) { echo '<tr><td> $row['player'] </td>'; echo '<td> . $row['city']</td>'; echo '<td> . $row['alliance']</td>'; echo '<td> . $row['x']</td>'; echo '<td> . $row['y']</td>'; echo '<td> . $row['other']</td>'; echo '<td> . $row['porter']</td>'; echo '<td> . $row['cons']</td>'; echo '<td> . $row['spy']</td>'; echo '<td> . $row['hbd']</td>'; echo '<td> . $row['mino']</td>'; echo '<td> . $row['lbm']</td>'; echo '<td> . $row['ssd']</td>'; echo '<td> . $row['bd']</td>'; echo '<td> . $row['at']</td>'; echo '<td> . $row['giant']</td>'; echo '<td> . $row['fm']</td>'; echo '<td> . $row['ft']</td>'; echo '<td> . $row['ogre']</td>'; echo '<td> . $row['banshee']</td></tr></table>'; } ?> Error shows up on line 35 but I'm not sure what I've done... Also, the xx's on my dbc statement were on purpose. Current error is: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/content/64/4940664/html/artofwar/browse.php on line 35, I can't figure out where the hell it wants a ;... Hi, So I'm trying to make a car rental project, and I need to show cars of a specific type, available in a specific period. I'm currently using this query: SELECT * FROM car, reservations WHERE car.car_id = reservations.car AND car.type = 1 AND reservations.reserved_from >= "2011-12-22" AND reservations.reserved_to <= "2012-12-20"; So this returns all the cars of the type 1 reserved in 2011-12-22 - 2012-12-20 period. So know that I know which cars are unavailable, how would you go ahead and pick out what's left? Hello, so i made table with Tags row which is Text(126).. and idea is to put text like "PHP, MySQL, Forum, Blog" etc etc...
I have problem searching tags my search code:
$requestedtag = $con->real_escape_string($_GET['tag']);
$sql = "SELECT * FROM images WHERE tags = '$requestedtag' ORDER BY ID DESC LIMIT 12";
$result = $con->query($sql);
if ($result->num_rows > 0)
{
while($row = $result->fetch_assoc()) {
$image_id = $row["ID"];
$image_title = $row["Title"];
$image_url = $row["ImgURL"];
$image_tags = $row["Tags"];
echo '<div class="col-3"><div class="thumb"><a href="/image.php?id='.$image_id.'" title="'.$image_title.'"><img src="'.$image_url.'"></img></a></div></div>';
}
}
But it doesnt return anything. I think the problem is that row["Tags"]; is divided by "," from tags and i should split it somehow but i dont really know what to do...
I have a multiple choice search page which asks visitors to select one or more choices before rendering the search results. It only works in items 1+2 are selected, or 1+2+3, or 1+2+3+4, or 1+2+3+4, etc... I want to make it so that any combination of choices work. Example: Items 1+6 or 2+3+5 or 1+4+6 etc... I have four hours to fix it before my client shows up and need help BADLY. Anyone out there who can help? Whoever can help me... you are the best and I will have a special prayer for you tonight. Here is the coding: <?php } else{ if(!empty($_GET['latt']) && !empty($_GET['long'])) { $condition="SELECT ( 3959 * acos( cos( radians( '".$_GET['latt']."' ) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians( '".$_GET['long']."' ) ) + sin( radians( '".$_GET['latt']."' ) ) * sin( radians( latitude ) ) ) ) as distance,id,membership FROM notary_members "; if(!empty($_GET['s'])) { $a=" where and"; } else { $a=" where"; } if($_GET['edoc']!='') { $condition.=$a." edoc= 'yes'"; } if($_GET['esign']!='') { if($_GET['edoc']!='') $condition.=" and esign= 'yes'"; else $condition.=$a." esign= 'yes'"; } if($_GET['laserprinter']!='') { if($_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and laserprinter= 'yes'"; else $condition.=" laserprinter= 'yes'"; } if($_GET['dualtray']!='') { if( $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and dualtray= 'yes'"; else $condition.=" dualtray= 'yes'"; } if($_GET['error_in']!='') { if($_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and error_in= 'yes'"; else $condition.=" error_in= 'yes'"; } if($_GET['bg_checked']!='') { if($_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and bg_checked= 'yes'"; else $condition.=" bg_checked= 'yes'"; } if($_GET['home_insp']!='') { if($_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and home_insp= 'yes'"; else $condition.=" home_insp= 'yes'"; } if($_GET['fingerprint']!='') { if($_GET['home_insp']!='' || $_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and fingerprint= 'yes'"; else $condition.=" fingerprint= 'yes'"; } if($_GET['hosp_signing']!='') { if($_GET['fingerprint']!='' || $_GET['home_insp']!='' || $_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and hosp_signing= 'yes'"; else $condition.=" hosp_signing= 'yes'"; } if($_GET['jail_signing']!='') { if($_GET['hosp_signing']!='' || $_GET['fingerprint']!='' || $_GET['home_insp']!='' || $_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and jail_signing= 'yes'"; else $condition.=" jail_signing= 'yes'"; } if($_GET['im_docs']!='') { if($_GET['jail_signing']!='' || $_GET['hosp_signing']!='' || $_GET['fingerprint']!='' || $_GET['home_insp']!='' || $_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='') $condition.=" and im_docs= 'yes'"; else $condition.=" im_docs= 'yes'"; } if($_GET['jail_signing']!='' || $_GET['hosp_signing']!='' || $_GET['fingerprint']!='' || $_GET['home_insp']!='' || $_GET['bg_checked']!='' || $_GET['error_in']!='' || $_GET['dualtray']!='' || $_GET['laserprinter']!='' || $_GET['esign']!='' || $_GET['edoc']!='' || $_GET['im_docs']!='') { $condition.=" and status=1 order by distance asc"; } else { $condition.=$a." status=1 order by distance asc"; } $sql=mysql_query($condition); $num=mysql_num_rows($sql); $getmile=explode(' ',$_GET['miles']); $c=0; while($getidres=mysql_fetch_array($sql)) { if($getidres['distance']<=$getmile[0]) { if($getidres['membership']=='premium') { $prefirst[$getidres['id']]=$getidres['distance']; $c++; } else { $basicfirst[$getidres['id']]=$getidres['distance']; $c++; } } } if(!empty($sql)) { if($c!=0) { if(!empty($prefirst) || !empty($basicfirst)) { ?> Not sure if this topic goes in here, it is related to PHP but also MySql, so if i'm on the wrong board sorry! What i'm trying to do is search for a keyword in 5 different tables and return the keyword ID from the table that its in The tables i'm trying to search are as follow location state county region continent The "location" table has all the locations i.e cities and each row has the following columns: id | continent_id | country_id | state_id | region_id | city_name The "state" table is set the the following: id | name "county" table : id | name "region" table: id | region and "continent" table id | name The way it works is the can search for any city or state or county or region or continent and ideally it should look into the five different tables and return the id of that table. So if the use searches for United States it will look for United States in all five tables, obviously it would find it in the "country" table so it should return that "id". The results are returned in "json format" Below is the code i have: Code: [Select] <? $input = $_GET['keyword']; $data = array(); /* In this query i'm attempting to search in all databases, but i'm not sure if i'm doing this right. I'm not getting any results so i know something is wrong just don't know how to write it. */ $query = mysql_query("SELECT * FROM locations JOIN states ON states.id = locations.state_id JOIN countries ON countries.id=locations.country_id JOIN regions ON regions.region_id = locations.region_id JOIN continents ON continents.id=locations.continent_id WHERE name LIKE '$input%' OR state LIKE '$input%' OR region LIKE '$input%' OR country LIKE '$input%' OR continent LIKE '$input%'") or die(mysql_error()); /* Here the values are added to to the $json array. The "value" should be the "id" from the table that the keyword matched. The 'name' Should be the name of the actual keyword. Again if they search for United States the "id" will come from the "countries" table and the "value" would come from the "countries" table as the name */ while ($row = mysql_fetch_assoc($query)) { $json = array(); $json['value'] = $row['id']; $json['name'] = $row['name']; $data[] = $json; } header("Content-type: application/json"); echo json_encode($data); ?> Any help would he be appreciated, i don't want people to do it for me, but rather just guide me a little bit. Hi. I have a database named 'Forums' with a table named 'forumlist'. The fields in the table are 'id', 'name', 'link', and 'keys'. I am trying to search the database in either the 'name' or 'keys' field and display the matches. I am getting this error: Parse error: syntax error, unexpected T_STRING, expecting ']' in C:\xampp\htdocs\search3.php on line 51 (The define fields) <html> <head> <title>My Forum Search</title> <style type="text/css"> table { background-color: #CCC } th { width: 150px; text-align: left; } </style> </head> <body> <h1>Forum Search</h1> <form method="post" action="search3.php"> <input type="hidden" name="submitted" value="true" /> <label>Search Category: <select name="category"> <option value="keys">Keyword</option> <option value="name">Name</option> </select> </label> <label><input type="text" name="criteria" /></label> <input type="submit" /> </form> <?php if (isset($_POST['submitted])) { DEFINE ('DB_USER', 'root'); DEFINE ('DB_PSWD', '******'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'Forums'); $dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME); $category = $_POST['category']; $criteria = $_POST['criteria']; $query = "SELECT * FROM forumlist WHERE $category LIKE '$criteria'"; $result = mysqli_query($dbcon, $query) or die ('Error retrieving data') echo "<table>"; echo "<tr> <th>Name</th><th>Link</th> </tr>"; while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['link']; echo "</td></tr>"; } echo "</table>"; } // end of main if state ?> </body> </html> Thanks in advance Hi, I'm am trying to use a search engine that search's for a username in the database and displays the information back. I have searched for a script but none of them has helped me. I have tried to use this without any luck: mysql_connect("localhost", "", "") or die(mysql_error()); mysql_select_db("dbsystem") or die(mysql_error()); $todo=$_POST['todo']; if(isset($todo) and $todo=="search"){ $search_text=$_POST['search_text']; $type=$_POST['type']; $search_text=ltrim($search_text); $search_text=rtrim($search_text); if($type<>"any"){ $query="select * from users where name = '$search_text'"; }else{ $kt=split(" ",$search_text);//Breaking the string to array of words // Now let us generate the sql while(list($key,$val)=each($kt)){ if($val<>" " and strlen($val) > 0){$q .= " name like '%$val%' or ";} }// end of while $q=substr($q,0,(strlen($q)-3)); // this will remove the last or from the string. $query="select * from users where $q "; } // end of if else based on type value echo $query; echo "<br><br>"; $nt=mysql_query($query); echo mysql_error(); while($row=mysql_fetch_array($nt)){ echo "$row[name]<br>"; } // End if form submitted }else{ echo "<form method='post' action=''><input type='hidden' name='todo' value='search' /> <input type='text' name='search_text' /><input type='submit' value='Search' /><br> <input type='radio' name='type' value='any' checked />Match any where <input type='radio' name='type' value='exact' />Exact Match </form> "; } I will be grateful if you can help with this. I'm using some code to create a select menu of a fieldname of data I have in a mysql database : Code: [Select] <?php mysql_connect('localhost' , 'dbname', 'password'); mysql_select_db('dbname'); $result=mysql_query("SELECT * FROM Persons"); if(mysql_num_rows($result)>0) { ?> <select name="Persons"> <?php while($rows=mysql_fetch_array($result)){ ?> <option value="<?php echo $rows['id']; ?>"> <?php echo $rows['FirstName']; ?></option> <?php } ?> </select> What I would like to do is to elevate this into a jump menu form so that if the user selects an item from my form they are taken to a results page showing the full row of data from the database. Basically a search form containing items from the database they can choose to see more details on. eg. In my example you select a persons name and then you are taken to a results page which displays the details of that person from the database. Problem is I don't know how to do this and have been trawling around for a couple of days to find a solution (sorry I'm new to php). I would appreciate some help or a working example would be great of : 1/ A working dynamic jump menu 2/ The page that would process the form 3/ The results page displaying the data I have selected. Thank you for your time... I want to have a search product feature, but I would like members to be able to search multiple fields in one go i.e. product_code, Product_name in one MySQL query. The thing is, members have to be logged on, so the query must also only show results relating to that specific member, via the session[member_ID], my current query for listing products for that specific member is : Code: [Select] $sql = "SELECT productId, productCode, image, name, price, stock_level FROM product_inventory WHERE memberr_ID = '" . $_SESSION['SESS_mem_ID'] . "; How would I change the above into a search query to search for productcode, productname and still only show results beloging to this member using the session data ? all help appreciated I have a simple (1 table, 14 fields) database. I'm using 9 separate search boxes from my search page : http://www.cinemaposter.com/database.html There are currently 162 entries. Hitting "submit" throws them all out, same with valid input into any box, with the exception of "country" - example : entering "F" gives 20results; "PL" gives 34 results. Here's my code (to simplify, I only went down to the 4th box ('country' - the last and only one giving correct results). I'm one sorry newbie.... please help ! Here's the code : ------------------ <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "http://www.w3.org/TR/1998/REC-html40-19980424/loose.dtd"> <html> <head> <title>Search Results</title> <meta name="generator" content="BBEdit 6.5"> <style type="text/css"> td { color: #000000; font-family: arial; font-size: 10px; } </style> </head> <body> <?php $link = mysql_connect("xxxx", "xxxx", "xxxx") or die("Could not connect"); ///print "Connected successfully<BR>"; mysql_select_db("xxxx") or die("Could not select database"); { $Author = $_POST['Author']; $sql = "SELECT * FROM posters WHERE Author LIKE '%$Author%'"; } { $Title = $_POST['Title']; $sql = "SELECT * FROM posters WHERE Title LIKE '%$Title%'"; } { $Originaltitle =$_POST['Originalitle']; $sql="SELECT * FROM posters WHERE Originaltitle LIKE '%$Originaltitle%'"; } { $Country =$_POST['Country']; $sql="SELECT * FROM posters WHERE Country LIKE '%$Country%'"; } $result = mysql_query ($sql) or die("Query failed : " . mysql_error()); $numrs = mysql_num_rows($result); if ($numrs > 0){ if ($numrs==1){ print ("<p><font face=\"Arial\"><b>There is $numrs record in our database.</B> <BR><BR>"); }elseif($numrs>1){ print ("<p><font face=\"Arial\"><b>There are $numrs records in our database. </B><BR><BR>");//Apparently, he wan't to be able to search again...</b><form method \"POST\" action=\"$PHP_SELF\">Search Again:"); } echo("<a href='javascript:history.go(-1)'>Go Back</a></p>"); print("<P align=center>Click on the column heading links to sort the results.<br><br />"); print ("<TABLE CELLPADDING=\"1\" CELLSPACING=\"2\" BORDER=\"0\">\n"); PRINT ("<TR BGCOLOR=\"#FFFF77\"> <TD COLSPAN=2><P align=center>Author</td> <TD><P align=center>Title</TD> <TD><P align=center>Original Title</TD> <TD><P align=center>Country</TD> <TD><P align=center>Director</TD> <TD><P align=center>Cast</TD> <TD><P align=center>Year</TD> <TD><P align=center>Original Year</TD> <TD><P align=center>Photo</TD> <TD><P align=center>Producer</TD> <TD><P align=center>Circulation</TD> <TD><P align=center>Size</TD> <TD><P align=center>Print type</TD> <TD><P align=center>Remarks</TD></TR>"); $z=1; while ($line = mysql_fetch_array($result)) { $num=$z++; //Note here that I'm assigning the found objects new names (indicated by lowercase) to that php wont get confused..... $auth=$line['Author']; $title=$line['Title']; $originaltitle=$line['Originaltitle']; $country=$line['Country']; $director=$line['Director']; $cast=$line['Cast']; $year=$line['Year']; $originalyear=$line['Originalyear']; $producer=$line['Producer']; $circ=$line['Circulation']; $size=$line['Size']; $print=$line['Print']; $remarks=$line['Remarks']; $pict=$line['Picture']; //Set the formatting $bgcolor = ($i++ & 1) ? '#d3d3d3' : '#c0c0c0'; PRINT("<TR BGCOLOR=$bgcolor> <TD>$num.</TD> <TD>$auth</td> <TD>$title</TD> <TD>$originaltitle</TD> <TD><P align=center>$country</TD> <TD>$director</TD> <TD>$cast</TD> <TD><P align=center>$year</TD> <TD><P align=center>$originalyear</TD><TD>"); // This is the test to find do we have a picture if ($pict) { // If there is a picture, display it: print ("<P align=center><a href='zdjecie.php?zdj= $pict&autor=$auth&opis=$title&orig=$originaltitle&rok=$year' target='_blank'> <img src=$pict height=\"150\"></a>"); } else { // We don't have a picture, so we print some text instead: print ("<P align=center>no picture available"); } print ("</TD><TD>$producer</TD> <TD><P align=center>$circ</TD> <TD>$size</TD> <TD>$print</TD> <TD>$remarks</TD></TR>"); } PRINT("</TABLE><BR><font face=\"Arial\"><a href=\"javascript:history.go(-1);\">Search Again</a></font>"); }else { print ("<font face=\"Arial\">Sorry, there were no results for your search. <a href=\"javascript:history.go(-1);\">Please try again</a></font>"); } //mysql_close($link); ?> </body> </html> Say a user puts in a support request, and for every request it generates a unqiue string, and enters it into the database. Ok, now say there is a text field, when the user enters their unique string and it finds a match, it displays the data along with it. How can I accomplish this? Im kind of new to mysql, but I know basic SQL. Would be great if somebody could point me in the right direction! Thanks This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=326703.0 |
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