Alidad Posted September 13, 2009 Share Posted September 13, 2009 Hi, i have wrote forms that allow me to upload images and save into the folder and send image file path name to the database, i have run it and test is works fine when save images into the folder but is not sending data to the mysql database. i'm not sure what went wrong but the result is not telling us the errors. it said it has send to database but the truth is not sending to database. This is mysql table name please create folder name images CREATE TABLE employees (name VARCHAR(30), email VARCHAR(30), phone VARCHAR(30), photo VARCHAR(30)) this index.php form page <form enctype="multipart/form-data" action="add.php" method="POST"> Name: <input type="text" name="name"><br> E-mail: <input type="text" name = "email"><br> Phone: <input type="text" name = "phone"><br> Photo: <input type="file" name="photo"><br> <input type="submit" value="Add"> </form> this is retrieve information from index.php page <?php //This is the directory where images will be saved $target = "images/"; $target = $target . basename( $_FILES['photo']['name']); //This gets all the other information from the form $name=$_POST['name']; $email=$_POST['email']; $phone=$_POST['phone']; $pic=($_FILES['photo']['name']); // Connects to your Database mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()) ; mysql_select_db("Database_Name") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')") ; //Writes the photo to the server if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> and this final step to view it. <?php // Connects to your Database mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()) ; mysql_select_db("Database_Name") or die(mysql_error()) ; //Retrieves data from MySQL $data = mysql_query("SELECT * FROM employees") or die(mysql_error()); //Puts it into an array while($info = mysql_fetch_array( $data )) { //Outputs the image and other data Echo "<img src=http://www.yoursite.com/images/".$info['photo'] ."> <br>"; Echo "<b>Name:</b> ".$info['name'] . "<br> "; Echo "<b>Email:</b> ".$info['email'] . " <br>"; Echo "<b>Phone:</b> ".$info['phone'] . " <hr>"; } ?> please help thanks. AM Link to comment https://forums.phpfreaks.com/topic/174102-need-help-with-image-path/ Share on other sites More sharing options...
cbolson Posted September 13, 2009 Share Posted September 13, 2009 Hi, Is it adding any data to the database (name, email etc.)? You should add some error checking to your code after a db query: mysql_query("INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')") or die("Error with insert.<br>".mysql_error()); To be able to further debug your code you need to make a variable of the insert : $insert="INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')"; mysql_query($insert) or die("Error with insert.<br>".mysql_error()); This way you can easily do echo $insert to check the insert data. Also, why do you have brackets in this line?: $pic=($_FILES['photo']['name']); They might be messing things up. Chris Link to comment https://forums.phpfreaks.com/topic/174102-need-help-with-image-path/#findComment-917746 Share on other sites More sharing options...
Aravinthan Posted September 13, 2009 Share Posted September 13, 2009 I think cbolson is right for the: $pic=($_FILES['photo']['name']); Just to make sur can you output $pic to see if its correct? Link to comment https://forums.phpfreaks.com/topic/174102-need-help-with-image-path/#findComment-917851 Share on other sites More sharing options...
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