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PHP - Php & Mysql: Using $_get['id'] To Query Matching Table Id
Hey guys having a bit of bother here, I don't understand why my PHP script isn't working.
So I will first show the code and then go over the issue I am having:
include('select-stock.php');
include('db-affinity/header-main.php'); ?>
<?php $carId = $_GET['id']; ?>
<?php
try {
$carview = $db->prepare("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive, FullRegistration FROM import WHERE FullRegistration = $carId");
} catch (Exception $e) {
echo "Error.";
exit;
}
$cardata = $carview->fetch(PDO::FETCH_ASSOC)
?>
<div class="container">
<div class="row">
<div class="col-md-12 col-sm-12">
<?php echo "$carId"; ?>
<?php echo mysql_errno($carview) ?>
<?php echo '<ul class="overwrite-btstrp-ul other-specs-ul h4-style">
<li>Mileage: '.number_format($cardata["Mileage"]).'</li>
<li>Engine size: '.$cardata["EngineSize"].'cc</li>
</ul>'
?>
</div>
</div>
</div>
<?php include('db-affinity/footer.php') ?>
So basically what I am trying to achieve from this code is giving my page dynamic content based on if the the ?id= of a URL matches a row of my 'FullRegistration' column.
So for example if I have a URL like this "www.cars.com/carview.php?id=NG61CWJ" I then want my script check if there is a row that has that value in the 'FullRegistration' column of my table and then echo out the results of certain columns of that row like this example currently in my code:
<?php echo '<ul class="overwrite-btstrp-ul other-specs-ul h4-style"> <li>Mileage: '.number_format($cardata["Mileage"]).'</li> <li>Engine size: '.$cardata["EngineSize"].'cc</li> </ul>' ?>In theory FROM import WHERE FullRegistration = $carIdshould make this happen however for some reason on my server when I use the script above I get nil results returned instead of the results of the row that matches the GET id I get: Mileage: 0 Engine size: cc I am aware my code is insecure at the moment however it isn't an issue at this moment in time. Any ideas why I might be getting nil results returned, my other queries to this table have worked flawlessly however I am having bother with this one, can you see anything in this code that might cause this issue? Here is the database connection file that is included at the top of the code block just in case this could be a bit of a problem:
<?php
include('database.php');
try {
$results = $db->query("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive, FullRegistration FROM import ORDER BY Make ASC");
} catch (Exception $e) {
echo "Error.";
exit;
}
///carousel-vehicle results
try {
$fourresults = $db->query("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive FROM import ORDER BY Make LIMIT 0, 4");
} catch (Exception $e) {
echo "Error.";
exit;
}
try {
$fourresultsone = $db->query("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive FROM import ORDER BY Make LIMIT 4, 4");
} catch (Exception $e) {
echo "Error.";
exit;
}
try {
$fourresultstwo = $db->query("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive FROM import ORDER BY Make LIMIT 8, 4");
} catch (Exception $e) {
echo "Error.";
exit;
}
try {
$makeFilter = $db->query("SELECT DISTINCT Make FROM import ORDER BY Make ASC");
} catch (Exception $e) {
echo "Error.";
exit;
}
try {
$modelFilter = $db->query("SELECT DISTINCT Model FROM import ORDER BY Make ASC");
} catch (Exception $e) {
echo "Error.";
exit;
}
?>
All of these queries are working flawlessly on the live site so the db connection is obviously working.Similar TutorialsHow can i use single quote for values? $qry='insert into tablename values('a','b');'; Hello there, i'm not sure if this goes to mysql, or this section, but since it's PHP based, i would say here. So, here's the thing, i have full set up table that reads from mysql database and display it as table. Everything is working just fine, even "order" buttons that i made, now i would like to have below the table some counts, for example: My table name "canonkickoff" has column named "Prevoz" which contain only Yes and No answers, i would like to see below the table how much "Yes" answers are inside the column. Also i have other column called "VelicinaMajice", which contain 6 different answers ( XS, S, M, L, XL, XXL ), i would like to see (again, below the table) how much of every has been answered. Like: There a XS = 6 S = 1 M = 2 L = 3 And 32 answered with Yes. Could anyone help me out? Here's the full php table: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <title>CanonKickOff 2012 Tabela.</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"> <link rel="stylesheet" href="table.css" type="text/css"> </head> <body style="margin: 0 0 0 0;"> <? include("passwd.php"); @$start = $_GET["start"]; if($start =='') $start =0; include("lib.php"); $link = mysql_connect($host,$username,$password); if (!$link) { die('Could not connect: ' . mysql_error()); } $db_selected = mysql_select_db($db, $link); if (!$db_selected) { die ("Can't use $db : " . mysql_error()); } //total number of records in the table $orderBy = array('Kompanija', 'id', 'ImePrezime', 'Email', 'Prevoz', 'VelicinaMajice', 'SlazemSe'); $order = 'id'; if (isset($_GET['orderBy']) && in_array($_GET['orderBy'], $orderBy)) { $order = $_GET['orderBy']; } $res = mysql_query("SELECT * from `$table` ORDER BY '.$order"); $res2 = mysql_query("SELECT * from `$table` "); @$rows = mysql_num_rows ($res2); $result = mysql_query("SELECT * from `$table` ORDER BY $order limit $start,40"); if (!$result) { die('Invalid query: ' . mysql_error()); } echo 'Sortiraj Po: <br>'; echo '<a href="?orderBy=id">ID:</a> '; echo '<a href="?orderBy=ImePrezime">Ime i Prezime:</a> '; echo '<a href="?orderBy=Kompanija">Kompanija:</a> '; echo '<a href="?orderBy=email">Email adresi:</a> '; echo '<a href="?orderBy=Prevoz">Prevoz:</a> '; echo '<a href="?orderBy=VelicinaMajice">Velicina majice:</a> '; echo '<a href="?orderBy=SlazemSe">Slazu se:</a> '; echo "<p align=center class = 'menu'> Ocitana tabela: $table </p>"; $cols = mysql_num_fields($result); $records = mysql_num_rows ($result); echo "<table align='center' width='1200' >"; echo "<tr bgcolor='BBCCDD' class='menu'>"; for ($i = 0; $i < $cols;$i++) { echo "<td align='center'>".mysql_field_name($result,$i)."</td>"; } echo "</tr>"; while ($row = mysql_fetch_array($result, MYSQL_NUM)) { echo "<tr bgcolor='F6F6F6' class='normal'>"; foreach ($row as $value) { echo "<td align='center'>".$value ."</td>"; } echo "</tr>"; } $end = $start + $records; echo "<tr align = 'center' bgcolor = 'BBCCDD' class='menu'><td colspan=$cols> $start do $end od ukupno: $rows </td></tr>"; echo "<tr align = 'center' class='mylink'><td colspan=$cols> "; if($start != 0) { $prev = $start - 40; echo "<a href='tabela.php?start=$prev'> Prethodna </a> "; } if($start<$rows-10) { $next = $start + 40; echo "<a href = 'tabela.php?start=$next'>Sledeca</a> "; } echo "</td></tr>"; echo "</table>"; ?> </body> </html> Hey guys, so I have this script:
<?php $databasehost = "localhost"; $databasename = "import"; $databasetable = "import"; $databaseusername="import"; $databasepassword = "password"; $fieldseparator = ","; $lineseparator = "\n"; $csvfile = "test.csv"; if(!file_exists($csvfile)) { die("File not found. Make sure you specified the correct path."); } try { $pdo = new PDO("mysql:host=$databasehost;dbname=$databasename", $databaseusername, $databasepassword, array( PDO::MYSQL_ATTR_LOCAL_INFILE => true, PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION ) ); } catch (PDOException $e) { die("database connection failed: ".$e->getMessage()); } $affectedRows = $pdo->exec(" LOAD DATA LOCAL INFILE ".$pdo->quote($csvfile)." REPLACE INTO TABLE `$databasetable` FIELDS TERMINATED BY ".$pdo->quote($fieldseparator)." LINES TERMINATED BY ".$pdo->quote($lineseparator)); echo "Loaded a total of $affectedRows records from this csv file.\n"; ?> This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=328883.0 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 the following code that searches my database and displays results in a table: $fields = array("field1", "field2", "field3") $cols = implode (', ', $fields); $result= mysql_query (" SELECT $cols FROM tablename WHERE ................... "); if (!$result) {die('Could not search database: ' . mysql_error());} if($result) { if(mysql_num_rows($result) == 0) { return "Sorry. No records found in the database"; } else { $table = "<table border='1' cellpadding='5' cellspacing='5'>"; while($arr = mysql_fetch_array($result, MYSQL_ASSOC)) { $table .= "\t\t<tr>\n"; foreach ($arr as $val_col) { $table .= "\t\t\t".'<td>'.$val_col.'</td>'."\n"; } $table .= "\t\t</tr>\n"; } $table .= "</table>"; echo $table; } mysql_free_result($result); } As you can see each of the MySQL table fields specified by $fields is displayed in a new column in the html table. I want to change this so that e.g. "field3" is displayed in a new row instead. So, instead of the html table looking like: | "field1-result1" | "field2-result1" | "field3-result1" | | "field1-result2" | "field2-result2" | "field3-result2" | | "field1-result3" | "field2-result3" | "field3-result3" | I want it to look like: | "field1-result1" | "field2-result1" | | "field3-result1" | | | "field1-result2" | "field2-result2" | | "field3-result2" | | | "field1-result3" | "field2-result3" | | "field3-result3" | | I guess this is quite straightforward, but I can't work it out! Pls help! Thanks. I have a photo album style gallery to build and i'm finding it dificult to list all the table names (these are names of photo albums) and then enter the data into a seperate query for each album name (these will change often so i cant keep updating the file as normal. this will then post all the data to the xml file and show the set of photos in the individual albums in a flash file. can anyone help me where im going wrong at all? <?php $dbname = 'cablard'; if (!mysql_connect('localhost', 'cablard', '')) { echo 'Could not connect to mysql'; exit; } $sql = "SHOW TABLES FROM $dbname"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row = mysql_fetch_row($result)) { echo "Table: {$row[0]}\n"; } mysql_free_result($result); $query = "SELECT * FROM photo ORDER BY id DESC"; $result2 = mysql_query ($query) or die ("Error in query: $query. ".mysql_error()); while ($row = mysql_fetch_array($result2)) { echo " <image> <date>".$row['date']."</date> <title>".$row['title']."</title> <desc>".$row['description']."</desc> <thumb>".$row['thumb']."</thumb> <img>".$row['image']."</img> </image> "; } ?> Thanks James Why exactly is the _get returning just nothing everytime? Code: [Select] $seshid = $_GET['finish']; // Little things in here ... ///Add lesson plan if(isset($_POST['addinc'])){ if(!empty($_POST['inc'])){ $inc = addslashes($_POST['inc']); mysql_query("INSERT INTO `group_sessions_lplan` (`group_sessions_lplan_id`, `group_session_id`, `plan`) VALUES ('', '$seshid', '".strip_tags($inc, "<a><b><strong><u><i><span>")."')"); header("Location: ?id=".$seshid."#I"); exit(); } } Not too sure myself, anyone know how I can get it to show its actual value? Here we go, I need to use a query where it uses a posted time value to compare if there are the same times on the posted date value. I want it so the user cant book the same time on the same day as someone before bascially. My input so far is this Code: [Select] <?php ob_start();?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Conforming XHTML 1.0 Strict Template</title> <link rel="stylesheet" type="text/css" href="style.css" /> <link type="text/css" href="ui-lightness/jquery-ui-1.8.16.custom.css" rel="Stylesheet" /> <script type="text/javascript" src="js/jquery-1.6.2.min.js"></script> <script type="text/javascript" src="js/jquery-ui-1.8.16.custom.min.js"></script> </head> <body> <form name="input" action="input.php" method="post"> Subject: <input type="text" name="subject" /> First Name: <input type="text" name="firstname" /> Surname: <input type="text" name="surname" /> Trainer: <input type="text" name="trainer" /> Email: <input type="text" name="email" /> Date: <input type="text" name="event_date" id="date" /> Time: <input type="text" name="event_time" id="time" /> <input type="submit" value="Submit" name="submit" /> </form> <script type="text/javascript"> $('#date').datepicker(); $('#time').timepicker({}); </script> <?php include_once("functions/database.php"); include_once("functions/number.php"); if (isset($_POST["submit"])) { echo $_POST['event_date']; echo mdy2mysql($_POST['event_date']); echo $_POST['event_time']; echo time2mysql($_POST['event_time']); $queryselect = "SELECT * FROM events LIKE '".$_POST['event_time']."'"; if ($queryselect == true) { echo "sorry this time is already booked"; } else { $query = "INSERT INTO events (subject, firstname, surname, trainer, email, event_date, event_time, status) VALUES('".$_POST["subject"]."', '".$_POST["firstname"]."', '".$_POST["surname"]."','".$_POST["trainer"]."','".$_POST["email"]."' ,'".mdy2mysql($_POST['event_date'])."','".time2mysql($_POST['event_time'])."', 'pending' ) "; $result = mysql_query($query, $db_link) or die(mysql_error().'cannot get results!'); header("Location: input.php"); } ?> can anyone help me ? very much appreciated. I have 3 tables w/ the following fields: Cities id | state_id | city_name States id | state_name New_cities id | state_id | city_name I am trying to write a script (and still working on it) that will INSERT new cities in cities WHERE New_cities.state_id = Cities.state_id. Obviously I am still in testing and results sort out, but honest to all hell, I might be going about this all wrong, and would appreciate any comments, criticism, assistance or feedback on what I am trying to do. <?php $query = "SELECT * FROM cities, states, new_cities"; if ($results = mysqli_query($cxn, $query)) { $row_cnt = mysqli_num_rows($results); echo $row_cnt . " Total Records in Query.<br /><br />"; if (mysqli_num_rows($results)) { while ($row = mysqli_fetch_array($results)) { if ($row['cities']['state_id'] == $row['new_cities']['new_state_id']) { $insert_city_query = "INSERT INTO cities id, new_state_id, city_name VALUES ('','$new_state_id','$city_name')" or mysqli_error(); } //if ($row['cities']['state_id'] == $row['new_cities']['new_state_id']) echo "$row[new_city_name]<br />"; } //while ($row = mysqli_fetch_array($results)) } //if (mysqli_num_rows($results)) } //if ($results = mysqli_query($cxn, $query)) ?> In querying my database, I'm Selecting nameFirst and nameLast, and it produces a name like Joe Smith. I'm trying to match a photo with the name. Right now I'm uploading photos into a folder naming the file (e.g. SmithJoe.jpg). For reasons that involve writers being able to upload and access photos, I'm trying to use an image plugin. When uploading photos, it strips capital letters, so SmithJoe,jpg goes in as smithjoe.jpg, and it's not matching my database query. Here is the code I'm working with that works quite well with this one exception: $query = 'SELECT * FROM wp_playerRank'; $results = mysql_query($query); while($line = mysql_fetch_assoc($results)) { if ($line['wpID'] == $wp_tagID) { echo '<div class="player">'; // Here is the code that produces the image. I need to get rid of capital letters for ease of use echo '<div><img src="/wp-content/uploads/' . $line['nameLast'] . $line['nameFirst'] . '.jpg"></div>'; echo '<div class="playerData">' . $line['height'] . ' '; if ($line['position'] == 'PG') {echo 'Point Guard';} elseif ($line['position'] == 'SG') {echo 'Shooting Guard';} elseif ($line['position'] == 'SF') {echo 'Small Forward';} elseif ($line['position'] == 'PF') {echo 'Power Forward';} elseif ($line['position'] == 'C') {echo 'Center';} echo '</div>'; echo '<div class="playerData">' . $line['hschool'] . '</div>'; echo '<div class="playerData">' . $line['summer'] . '</div>'; echo '<div class="playerData">Class of ' .$line['year'] . '</div>'; if ($line['committed'] == 'y') { echo '<div> <br>Committed to <strong>'. $line['college'] . '</strong></div> ';} } } This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=351561.0 Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> A few months ago, and a good amount of time before that, I had people telling me to use isset() instead of performing to see if the variable is empty, such as: !$_GET[''] I know the differences in the function and what they do, but when could isset() be used in a situation where it's better/more efficient then: !$_GET[''] I do use isset(), though. I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas?
create table mimi (mimiId int(11) not null, mimiBody varchar(255) );
<?php
//connecting to database
include_once ('conn.php');
$sql ="SELECT mimiId, mimiBody FROM mimi";
$result = mysqli_query($conn, $sql );
$mimi = mysqli_fetch_assoc($result);
$mimiId ='<span>No: '.$mimi['mimiId'].'</span>';
$mimiBody ='<p class="leading text-justify">'.$mimi['mimiBody'].'</p>';
?>
//what is next?
i want to download pdf or text document after clicking button or link how to do that Hello, I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints? Thanks in advance, Vero Hello everyone, Sorry if this has been answered but if it has I can't find it anywhere. So, from the begining then. Lets say I had a member table and in it I wanted to store what their top 3 interests are. Their$ row has all the usual things to identify them userID and password etc.. and I had a further 3 columns which were labled top3_1 top3_2 & top3_3 to put each of their interests in from a post form. If instead I wanted to store this data as a PHP Array instead (using 1 column instead of 3) is there a way to store it as readable data when you open the PHPmyadmin? At the moment all it says is array and when I call it back to the browser (say on a page where they could review and update their interests) it displays 'a' as top3_01 'r' as top3_02 and 'r' as top3_03 (in each putting what would be 'array' as it appears in the table if there were 5 results. Does anyone know what I mean? For example - If we had a form which collected the top 3 interests to put in a table called users, Code: [Select] <form action="back_to_same_page_for_processing.php" method="post" enctype="multipart/form-data"> <input name="top3_01" type="text" value="enter interest number 1 here" /> <input name="top3_02" type="text" value="enter interest number 2 here" /> <input name="top3_03" type="text" value="enter interest number 3 here" /> <input type="submit" name="update_button" value=" Save and Update! " /> </form> // If my quick code example for this form is not correct dont worry its not the point im getting at :) And they put 'bowling' in top3_01, 'running' in top3_02 and 'diving' in top3_03 and we catch that on the same page with some PHP at the top --> Code: [Select] if (isset($_POST)['update_button']) { $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars $top3_02 = $_POST['top3_02']; // i.e, 'running' $top3_03 = $_POST['top3_03']; // i.e, 'diving' With me so far? If I had a table which had 3 columns (1 for each interest) I could put something like - Code: [Select] include('connect_msql.php'); mysql_query("Select * FROM users WHERE id='$id' AND blah blah blah"); mysql_query("UPDATE users SET top3_01='$top3_01', top3_02='$top3_02', top3_03='$top3_03' WHERE id='$id'"); And hopefully if ive got it right, it will put them each in their own little column. Easy enough huh? But heres the thing, I want to put all these into an array to be stored in the 1 column (say called 'top3') and whats more have them clearly readable in PHPmyadmin and editable from there yet still be able to be called back an rendered on page when requested. Continuing the example then, assuming ive changed the table for the 'top3' column instead of individual colums, I could put something like this - Code: [Select] if (isset($_POST)['update_button']) { $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars $top3_02 = $_POST['top3_02']; // i.e, 'running' $top3_03 = $_POST['top3_03']; // i.e, 'diving' $top3_array = array($top3_01,$top3_02,$top3_03); include('connect_msql.php'); mysql_query("UPDATE members SET top3='$top3_array' WHERE id='$id' AND blah blah blah"); But it will appear in the column as 'Array' and when its called for using a query it will render the literal string. a r r in each field instead. Now I know you can use the 'serialize()' & 'unserialize()' funtcions but it makes the entry in the database practically unreadable. Is there a way to make it readable and editable without having to create a content management system? If so please let me know and I'll be your friend forever, lol, ok maybe not but I'd really appreciate the help anyways. The other thing is, If you can do this or something like it, how am I to add entries to that array to go back into the data base? I hope ive explained myself enough here, but if not say so and I'll have another go. Thanks very much people, L-PLate (P.s if I sort this out on my own ill post it all here) Hello! Please help... I am trying to use the script below to get results from a mysql database based on a query of the form fields (the names of which are displayed near the top of the script as POST items) When a location or age etc. is entered into the form, I want the script to search for records which meet those criteria. At the moment the script works but only does so if all the values are entered to match what is in the database. e.g. if the location england and the age 22 was entered into the form, and that matched the value in the database, then at the moment, the script will display the result, but if only the location is entered in the form without any value for age/genre etc. then no results are displayed. Any help would be very welcome as I have search high and low for a solution on google... which doesn't seem to exist... I'm not that experienced with php/mysql but am learning on the job so any helpful prompts as to terms etc. would help! Thanks! Lewis <?php if($_POST) { $searchage = $_POST['searchage']; $searchlocation = $_POST['searchlocation']; $searchgenre = $_POST['searchgenre']; $searchinstrument = $_POST['searchinstrument']; $searchexperience = $_POST['searchexperience']; // Connects to your Database mysql_connect("localhost", "user", "pass") or die(mysql_error()); mysql_select_db("DB") or die(mysql_error()); $query = mysql_query("SELECT * FROM table_user WHERE userage = '".$searchage."' AND userlocation = '".$searchlocation."' AND usergenre = '".$searchgenre."' AND userinstrument = '".$searchinstrument."' AND userexperience = '".$searchexperience."'") or die(mysql_error()); $num = mysql_num_rows($query); echo "$num results found!<br>"; while($result = mysql_fetch_assoc($query)) { $username = $result['username']; $useremail = $result['useremail']; $userage = $result['userage']; $userlocation = $result['userlocation']; $usergenre = $result['usergenre']; $userinstrument = $result['userinstrument']; $userexperience = $result['userexperience']; $userbiography = $result['userbiography']; echo " Name: $username<br> Email: $useremail<br> Age: $userage<br> Location: $userlocation<br> Gen $usergenre<br> Instrument: $userinstrument<br> Experience: $userexperience<br> Biography: $userbiography<br><br> "; } } ?> |
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