android app connect to mysql with PHP
发布于2023-05-25 21:33 阅读(691) 评论(0) 点赞(26) 收藏(2)
I have no idea of php but the code should be something like this `
$con = mysqli_connect($host_name, $user_name, $user_pass, $db_name);
if($con)
{
$image = $_POST ["image"];
$name = $_POST ["name"];
$sql = "insert into imageinfo(name) values('$name')";
$upload_path = "uploads/$name.jpg";
if(mysqli_query($con,$sql))
{
file_put_contents($upload_path,base64_decode($image));
echo json_encode(array('response'=>'Image Upload Successfully'));
}
else
{
echo json_encode(array('response'=>'Image upload failed1'));
}
}
else
{
echo json_encode(array('response'=>'Image Upload Failed2'));
}
mysqli_close($con);
?>`
I'm getting an error of unidentified objets image and name from ($image = $_POST ["image"]..). If I use if(isset) I get the below response:
Image upload failed2
解决方案
Remove the space
$image = $_POST["image"];
$name = $_POST["name"];
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作者：黑洞官方问答小能手
链接：http://www.phpheidong.com/blog/article/546148/702f72b29f9a2850bb5e/
来源：php黑洞网
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