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pezzie

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  1. You are god. Thank you very much. i dnt know how to thank you mate. thank you
  2. any help. i still have not figured out what is wrong with that dam code of mine.
  3. Thanks for that. [color=red]index.php[/color] <?php session_start(); if (!$_SESSION["valid_user"]) { // User not logged in, redirect to login page Header("Location: login.php"); } // Display Member information echo "<p>User ID: " . $_SESSION["valid_id"]; echo "<p>Username: " . $_SESSION["valid_user"]; echo "<p>Logged in: " . date("m/d/Y", $_SESSION["valid_time"]); // Display Delete, Edit and add article links echo "<p><a href=\"articles.php\">VIEW all Articles</a></p>"; echo "<p><a href=\"add.php\">Add New Article</a></p>"; echo"<br>"; // Display logout link echo "<p><a href=\"logout.php\">Click here to logout!</a></p>"; ?> [color=red]edit.php[/color] <? // Connect database include("dbConfig.php"); // Get all records in all columns from table and put it in $result. $result=mysql_query("select * from articles ORDER BY id"); /*Split records in $result by table rows and put them in $row. Make it looping by while statement. */ while($row=mysql_fetch_assoc($result)){ // Output echo "ID : {$row['id']} <br/>"; echo "Title : {$row['title']} <br/>"; echo "Content : {$row['content']} <br/>"; echo "Filename : {$row['filename']} <hr>"; // Add a link with a parameter(id) and it's value. echo '<a href="update.php?id='.$row['id'].'">Update</a>'; echo "<HR>"; } mysql_close(); ?> [color=red]update.php[/color] <?php // Connect database. include("dbConfig.php"); if($_POST['Submit']){ // Get parameters from form. $id=$_POST['id']; $title=$_POST['title']; $content=$_POST['content']; $filename=$_POST['filename']; //mysql_query("update articles set title='$title', content='$content', filename='$filename' where id='$id'"); $q="UPDATE `articles` SET title='$title', content='$content', filename='$filename' where id='$id'"; $r = mysql_query($q); if ( !$r )   {     echo "ERROR Please try again"; } else { header("location:edit.php"); exit; } } // Get id parameter (GET method) from edit.php $id=$_GET['id']; // Get records in all columns from table where column id equal in $id and put it in $result. $result=mysql_query("select * from articles where id='$id'"); // Split records in $result by table rows and put them in $row. $row=mysql_fetch_assoc($result); echo $id; ?> <html> <body> <form id="form1" name="form1" method="post" action="<? echo $PHP_SELF; ?>"> <p>Name : <input name="name" type="text" id="title" value="<? echo $row['title']; ?>"/> <br /> Email : <input name="email" type="text" id="content" value="<? echo $row['content']; ?>"/> <br /> Tel : <input name="tel" type="text" id="filename" value="<? echo $row['filename']; ?>"/> </p> <p> <input type="submit" name="Submit" value="Submit" /> </p> </form> </body> </html>
  4. checked everything and no luck. This is a joke man. no one knows what the hell is wrong with that code. the code is getting on my nerves.  >:( .
  5. after i changed the code to the above reply, i got this. [color=red]Parse error: parse error, unexpected $ in /var/www/vhosts/acreinternet.co.uk/subdomains/stage2/httpdocs/new/update.php on line 65[/color] line 65 is the HTML TAG </body>
  6. When i echo out my $result, the output is [color=red]Resource id #4[/color]
  7. now it goes to page edit, BUT the data is the same.
  8. Thank you for your replies. everything is fine now and the text boxes appear with the data inside them BUT once i click on submit, nothing happens . EXAMPLE: if i change name from DAVID to JOHN and click on update, the page refreshes and the name changes to DAVID again.
  9. Thank for the reply. it is now giving me an error. Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/acreinternet.co.uk/subdomains/stage2/httpdocs/new/update.php on line 23
  10. Hi i want to get data from mySQL and wish to delete or update it. The delete is fine but when it comes to update, nothing happens at all. i want the data to be present in the textboxes so that the user does not have to type the whole thing in again.  [color=red]PLEASE PLEASE help. i need it by tomorrow.[/color] Thank you. this is my code: [color=red]edit.php[/color] <? // Connect database include("dbConfig.php"); // Get all records in all columns from table and put it in $result. $result=mysql_query("select * from articles ORDER BY id"); /*Split records in $result by table rows and put them in $row. Make it looping by while statement. */ while($row=mysql_fetch_assoc($result)){ // Output echo "ID : {$row['id']} <br/>"; echo "Title : {$row['title']} <br/>"; echo "Content : {$row['content']} <br/>"; echo "Filename : {$row['filename']} <hr>"; // Add a link with a parameter(id) and it's value. echo '<a href="update.php?id='.$row['id'].'">Update</a>'; echo "<HR>"; } mysql_close(); ?> //------------------------------------------------------------------------------------------ [color=red]update.php[/color] <? // Connect database. include("dbConfig.php"); if($Submit){ // Get parameters from form. $id=$_POST['id']; $title=$_POST['title']; $content=$_POST['content']; $filename=$_POST['filename']; mysql_query("update articles set title='$title', content='$content', filename='$filename' where id='$id'"); header("location:edit.php"); // Get id parameter (GET method) from edit.php $id=$_GET['id']; // Get records in all columns from table where column id equal in $id and put it in $result. //$result=mysql_query("select * from articles where id='$id'"); // Split records in $result by table rows and put them in $row. $row=mysql_fetch_assoc($result); } ?> <html> <body> <form id="form1" name="form1" method="post" action="<? echo $PHP_SELF; ?>"> <p>Name : <input name="name" type="text" id="title" value="<? echo $row['title']; ?>"/> <br /> Email : <input name="email" type="text" id="content" value="<? echo $row['content']; ?>"/> <br /> Tel : <input name="tel" type="text" id="filename" value="<? echo $row['filename']; ?>"/> </p> <p> <input type="submit" name="Submit" value="Submit" /> </p> </form> </body> </html>
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