php代码在数据库中插入数据失败[重复]

Question

你好，我写了一个php代码，从android接收文件，保存文件，并将url插入数据库。它保存的pdf完美，但不插入任何东西在数据库和给我的错误失败。我有什么办法可以修好它，怎么修好它？？

这是我的php代码

<?php
require "conn.php";


$pdf= $_FILES['file']['name'];
$tempName= $_FILES['file']['tmp_name'];   
$folder="uploadedFiles/";
$url = "https://mydomain/uploadedFiles/".$pdf;    





if($conn){

    if(move_uploaded_file($tempName,$folder.$pdf)){
        $sql_upload = "INSERT INTO pmfiles (`file`) VALUES('$url')";

        $query = mysqli_query($conn,$sql_upload); 
        if ($query){
            echo json_encode(array( "status" => "true","message" => "Successfully Uploaded!") );
                             
        } else {
            echo json_encode(array( "status" => "false","message" => "Failed!") ); //this is the error I get
                     
        } 
    
    }              
} else {
    echo json_encode(array( "status" => "false","message" => "Connection Error") );
}



?>

Answer 1

你的问题是你忘了一个撇号。

但这里有一个更好的代码请使用准备好的语句

$pdf = $_FILES['file']['name'];
$tempName = $_FILES['file']['tmp_name'];
$folder = "uploadedFiles/";
$url = "https://mydomain/uploadedFiles/" . $pdf;


if (!move_uploaded_file($tempName, $folder . $pdf)):
    echo json_encode(array("status" => "false", "message" => "Invalid file or wasn't able to save the file"));
    return;
endif;

$prepare = mysqli_prepare($conn, "INSERT INTO pmfiles (`file`) VALUES(?)");

mysqli_stmt_bind_param($prepare, "s", $url);

if (!mysqli_stmt_execute($prepare)):
    echo json_encode(array("status" => "false", "message" => "Failed!"));
    return;
endif;

echo json_encode(array("status" => "true", "message" => "Successfully Uploaded!", "data" => $url));