Upload a file Using PHP and JavaScript
Question
I selected a image using:
<input type="file" id="pimg" name="pimg" accept='image/*'/>
My javascript code:
p_img =document.getElementById("pimg").value;
param= 'pn='+p_img;
xmlhttp.open("GET","add_prod.php?"+param,false);
xmlhttp.send();
My php code:
p_img=$_GET['img'];
$con = mysqli_connect('localhost', 'admin', 'admin', 'products');
$sql="INSERT INTO prod (img) VALUES ('$p_img')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
This will store only the name of the file. But i want to copy the file from pc to directory. Its necessary to use Javascript as I'm using complete add product to pass values using AJAX
4 answers
solution1
0 2013-09-09 12:38:42
Create index.html file
<!DOCTYPE html>
<html>
<body>
<form action="upload_file.php" method="post" enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>
Create Php file upload_file.php
<?php
if ($_FILES["file"]["error"] > 0) {
echo "Error: " . $_FILES["file"]["error"] . "<br>";
} else {
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Stored in: " . $_FILES["file"]["tmp_name"];
}
?>
solution2
0 2013-09-09 12:40:33
Use POST (form) to send data to the php file.
With $_FILES["pimg"]["tmp_name"] you can move the uploaded file (with the php function move_uploaded_file to you webserver.
Link to PHP Function http://php.net/manual/de/function.move-uploaded-file.php
solution3
0 2016-02-17 14:13:48
You can upload the file without page refresh using simple JavaScript and PHP. Using JavaScript, the file would be passed to the PHP file and using move_uploaded_file() function file would be uploaded to the server.
The live demo and source code would found from here - Upload file using JavaScript and PHP
solution4
0 2017-11-02 06:13:20
xmlhttp.open("GET","add_prod.php?"+param,false);
我认为 open 方法的参数必须包含 true。
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