2天我试图找出问题的原因。但是我根本找不到。它没有显示“DOESNT SHOW ME”,它返回else语句。它打印了很多信息。请忽略用户名,product_id,我正在测试这就是为什么我认为他们感谢你帮助谁
<?php
include("config.php");
if($conn)
{
$product_id = "39";
$username = "admin";
$input = $_POST['comment'];
$subject = $_POST['subject'];
$type = $_POST['type'];
$path = "img/products/";
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");
$name = $_FILES['photoimg']['name'];
$size = $_FILES['photoimg']['size'];
if(strlen($name))
{
list($txt, $ext) = explode(".", $name);
if(in_array($ext,$valid_formats))
{
if($size<(1024*1024)) // Image size max 1 MB
{
$actual_image_name = time()."-".$username."-".$subject.".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
$upload_like= $path.basename($actual_image_name);
$done = move_uploaded_file($tmp,$upload_like);
if($done){
echo "DOESNT SHOW ME :(";
//$insert_image = "UPDATE post SET product_img = '$upload_like' WHERE username='$username' AND id='$product_id";
//$run_image = $conn -> query($insert_image);
$insert_sql = "INSERT INTO post (username,subject,user_post,time_added,type,product_img) VALUES ('$username','$subject','$post', CURRENT_TIMESTAMP , '$type','$upload_like')";
$run_sql = $conn -> query($insert_sql);
if($run_sql)
{
echo "DONE";
}
else
{
echo "Something crazy happened.OMG!";
}
}
else
echo "failed". "<br>" . PHP_EOL;
echo "Actual Image Name: " .$actual_image_name . "<br>" . PHP_EOL;
echo "Path To: ".$path. "<br>" . PHP_EOL;
echo "Temp: " .$tmp. "<br>" . PHP_EOL;
echo "Last version: ".$upload_like. "<br>" . PHP_EOL;
echo "Subject: ".$subject. "<br>" . PHP_EOL;
echo "Message: ".$input. "<br>" . PHP_EOL;
echo "Type: ".$type. "<br>" . PHP_EOL;
}
else
echo "Image file size max 1 MB";
}
else
echo "Invalid file format..";
}
else
echo "Please select image..!";
}
?>