The below is my source code where I want to use the variable $r1 which is at the bottom php script in another php script which is on top on the same page. I need a simple solution which solves this problem. I want to use that variable in the update query which is present in the code.
下面是我的源代码,我想使用变量$ r1,它位于同一页面上的另一个PHP脚本的底部php脚本中。我需要一个简单的解决方案来解决这个问题。我想在代码中出现的更新查询中使用该变量。
<?php
$con=mysql_connect("localhost","root","") or die("could not connect to db");
mysql_select_db("test");
$valid_formats = array("jpg", "png", "gif", "zip", "bmp","MP4","3GP");
$max_file_size = 1024*10000; //100 kb
//$path = "uploads/"; // Upload directory
$count = 0;
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST"){
// Loop $_FILES to execute all files
foreach ($_FILES['files']['name'] as $f => $name) {
if ($_FILES['files']['error'][$f] == 4) {
continue; // Skip file if any error found
}
if ($_FILES['files']['error'][$f] == 0) {
if ($_FILES['files']['size'][$f] > $max_file_size) {
$message[] = "$name is too large!.";
continue; // Skip large files
}
elseif( ! in_array(pathinfo($name, PATHINFO_EXTENSION), $valid_formats) ){
$message[] = "$name is not a valid format";
continue; // Skip invalid file formats
}
else{ // No error found! Move uploaded files
move_uploaded_file($_FILES["files"]["tmp_name"][$f],"uploads/" . $_FILES["files"]["name"][$f]);
// Number of successfully uploaded files
$file="uploads/".$_FILES["files"]["name"][$f];
/* $sql = "Update media set path = '$file' where username = '$r1'";
if (!mysql_query($sql))
{
die('Error: ' . mysql_error());
}*/
}
}
}
}
$sql="select * from media";
$query=mysql_query($sql);
while($row=mysql_fetch_array($query))
{
$image=$row[2];
if($image!='')
{
echo "<img src='".$row['path']."' width='175' height='200' />";
}
}
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<title>Multiple File Upload with PHP - Demo</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div class="wrap">
<h1><a>Multiple File Upload with PHP</a></h1>
<?php
# error messages
if (isset($message)) {
foreach ($message as $msg) {
printf("<p class='status'>%s</p></ br>\n", $msg);
}
}
# success message
if($count !=0){
printf("<p class='status'>%d files added successfully!</p>\n", $count);
}
?>
<p>Max file size 100kb, Valid formats jpg, png, gif</p>
<br />
<br />
<!-- Multiple file upload html form-->
<form action="" method="post" enctype="multipart/form-data">
<span style="font-size:12pt;">Channel: </span> <select name="channels">
<option value="">Channel</option>
<?php
$sql_query="SELECT * FROM media";
$result1 = mysql_query($sql_query) or die(mysql_error());
while($data=mysql_fetch_row($result1))
{
?>
<option value="<?php $r1 = $data[0]; echo $r1;?>" ><?php $r1 = $data[1]; echo $r1 ?></option>
<?php
}
?>
</select>
<input type="file" name="files[]" id="image" multiple="multiple">
<input type="submit" value="Upload">
</form>
</div>
</body>
</html>
1 个解决方案
#1
0
If you want to have the value of the <select name="channels">
you can use this
如果你想拥有
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST"){
$r1 = $_POST["channels"];
//your code here
}
Hope this help.
希望这有帮助。
#1
0
If you want to have the value of the <select name="channels">
you can use this
如果你想拥有
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST"){
$r1 = $_POST["channels"];
//your code here
}
Hope this help.
希望这有帮助。