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Hi, there may be error in the following code: [CODE] $managerID=($SESSION["id"]); $manager=($SESSION["manager"]); $password=($SESSION["password"]);[/CODE] [U][B]Replace above three line instead of below three lines[/B][/U] [CODE]$managerID=($_SESSION["id"]); $manager=($_SESSION["manager"]); $password=($_SESSION["password"]); [/CODE] ----- [CODE]$sql=mysql_query("SELECT * FROM admin WHERE id='$managerID' AND username='$manager' AND password='$password' LIMIT 1");[/CODE]
if (date('H') >= 16:00 ){ if($_POST['submit']=="Save") { $name = $_POST['name']; $age = $_POST['age']; $addr = $_POST['addr']; if(mysql_query("insert into file(name,age,address) values ('$name','$age','$addr')")) echo "Record Save"; else echo "Error in Insert query"; } }else{ echo "Time exceeded, You cannot insert into database. Try, again."; }
Hi, try the below code, you will got it. [CODE]<?PHP $query = mysql_query("SELECT * FROM call_details WHERE !ISNULL(call_project_name)"); echo "<select name='detailr'>"; while($row=mysql_fetch_array($query)) { if ($row['call_project_name'] != " "){ echo "<option value =". $row['call_id'].">". $row['call_project_name']."</option>"; } } echo "</select>"; ?>[/CODE]
[CODE]<html> <form method="post" action="test.php"> User Name : <input name="UserName" type="text" id="UserName" value="User Name" size="20" maxlength="30" /> <input type = "submit" name = "sumbit" value = "Submit"> </form> </html> <?php mysql_connect("Localhost","username","password") or die("Failure to communicate"); mysql_select_db("mydb") or die("Could not connect to Database"); if($_POST['submit'] == 'sumbit'){ if(!$_POST['UserName']){ echo "Enter User Name";}} else{ …
// Get an image based on id; $id = $_GET['id']; $image = "Select image_field_name from $table_name where id = '$id' "; $result = mysql_query($image); //create an image header('Content-type: image/jpeg'); $samp_image = imagecreatefromjpeg($result['image_field_name']); //display an image; imagejpeg($samp_image); // Free up memory imagedestroy($im);
The End.