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Hi There, Im really really pulling my hair out here, and was hoping you guys could help a really poor coder! basically, I have a table with a list of contracts in it, and a table with a list of applications for each contract. I can succesfully populate a list …

You are inserting 6 values into the table with only 5 columns..I think you missed out uname in the insert statement.

I have a select box that looks up values from my database, and then runs a jquery function to display the results, function is working fine, but some of the values in the database have spaces i.e "BT Global" the results are returning everything that starts with "BT" any help …

I have a curious issue with Twitter bootstrap, I include the bootstrap.js , but my popovers and tooltops dont work? `<button type="button" class="btn btn-lg btn-danger" data-toggle="popover" title="Popover title" data-content="And here's some amazing content. It's very engaging. Right?">Click to toggle popover</button>` All my other js stuff works fine, I have a …

In the process of converting an old web application I did from MYSQL to MYSQLi and am a bit lost with echoing a result, This code works in the old app with Mysql: <textarea name="dump" style="width:300px; height:400px;"> <?php include ("../connect.php"); $content = mysql_query("SELECT $field FROM customer WHERE cusid = '$cusid'") …

I import a spreadsheet (csv) into my sqli database (keeps a track of all our firewalls) there are 10 excel columns , one for each connected network on the firewall. i would like all these columns to be imported as one entry in the database, so 1 database field would …

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp\www\firewall\php\allresults.php on line 3 The annoying thing is, I have used this multiple times on different applications, I dont know why it think the result is boolean? the exact same code works on a different database, I just cant …

I want to have a supplier table, with ID, NAME, CONTACT, PHONE, etc, In my main table, a record can have more than one of these suppliers, what is the best way to select multiple rows from the supplier table if they relate to a record in the main table? …

`<a href="#" onclick='contacts($code)'>Contact Details</a>` This runs a jquery code with the value $code. in Firebug I can see it gets the correct rowset value, the jquery is below: function contacts(code){ jQuery.ajaxSetup({async:true}); jQuery.post("php/contact.php", {code: code}, function(data) { $('#results').hide(); $('#results').fadeIn(2000); $('#results').show() .html(data); }); } it is supposed to post this value to …

Hi, I use a query which is identical throughtout my application, only thing that differs is the select part of the query, so for example: mysql_select_db("mydatabase", $con); $result4 = mysql_query("SELECT * FROM nap where Exchange LIKE '$majorvalue%'"); while($row = mysql_fetch_array($result4)) { $psba = $row['Site']; $status = $row['Status']; $spectrum = $row['InSpectrum']; …

Hi, I have been able to write to a text file, then another page can read from that file, all good.. Problem is I'm using multiple txt files just to set various variables. There is a reason I'm not using a database. Can I somehow, write to a file after …

Hi, I have a large monitor at work on the wall, and I want to display some simple information, I have set up WAMP on my desk PC, I want to enter information on my desk PC and submit that data, and it displays on the main screen, I have …

Would this work maybe? echo "<div>" .$abc[ID] $id."</div>";

I have the below code that grabs all the data I need, but as there are duplicate names in each row, I only want to display this once as a heading, and then display the array. <?php echo "<h1>Results Summary</h1>"; //TABLE RESULTS $psba = $_POST['psba']; $con = mysql_connect("localhost","root",""); mysql_select_db('project', $con) …

Most annoyingly I cant get this to work, I made a sample table where there will be duplicate entries in some of the fields, and I need to display then all. table looks a bit like this id county city phonenumber the id is unique as is the phonenumber, but …

Hi There, I have written a very simple notice board page, for displaying information to the team where I work. I want to have an admin.html page where I can enter details in fields and select from drop downs, then hit submit and it populates a page called "notice.html" is …

I have no idea why this query doesnt work? the WHERE part will work if I put the value in of the rowset, but I can echo out the variable $faultid and its the same? it executes the query but doesnt change the record? all the variables are declared before …

Hi, I have a simple query here, one of my database fields is boolean, 1 or 0, when I echo out the value I want it to be yes or no, I did this, but it doesnt work, $value = .$row[booleanresult]; if ($value ="1"){ $value = "YES" } else { …

You can just echo it, `echo "<img src=". $image ." />";`

I have a script that uploads files to a folder and it works lovely when im testing on my XAMPP box, if someone else tries to upload it says Warning: copy(C:/xampp/htdocs/data/mottramg/AEGON.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs\training\upload.php on line 27 Could not copy file. now …

I have a "New Employee" script, that when you enter the details it creates a folder according to their username, using mkdir, this all works fine, I now have managed to import 200 entries into my DB from a spreadsheet, but I need to create a folder for all the …

I have a simple form that uses a bit of jquery, <form action="" method="post" enctype="multipart/form-data"> <label for="file">Upload File:</label> <input type="file" name="userfile" id="userfile" /> <input type="button" value="Submit" onclick="uploadfile()"/> </form> the uploadfile() in my .js file looks like this: function uploadfile(){ var userfile = document.getElementById('userfile').value; var username = document.getElementById('username').value; $.post("php/upload.php", {userfile: userfile,username: …

I have this in my index.php include ('php/config.php'); $connection = new db(); $connection->connect(); the class (in a file called config.php) looks like this: <?php //CONECTION CLASS class db { var $host="localhost"; var $username="root"; Var $password=""; var $database="training"; public function connect() { mysql_connect($this->host,$this->username,$this->password); mysql_select_db($this->database)or die(mysql_error()); $sql = "SHOW TABLES"; $result = …