Question

我有以下代码将图像的路径存储到数据库中。

文件路径未存储到数据库中。虽然图像名称是？

为什么会这样？

  <?php

  session_start();  
 error_reporting(E_ERROR | E_PARSE);


// escape variables for security
$ImageName = mysqli_real_escape_string($con, $_POST['imageName']);


  //getting the filename of the image file.
 $file_name = $_FILES['fileToUpload']['name'];


   $sql="UPDATE images SET imageLoc = '$file_name', imageName = '$ImageName'
    WHERE Username = '".$_SESSION["Username"]."'";
//$query = mysqli_query($con,$sql);

     if (!mysqli_query($con,$sql)){

  echo "error";

   }

 echo "yess";



 mysqli_close($con);
 ?>

数据库连接正常。

由于

XHTML部分：

  Select image to upload:

    <input type="file" name="fileToUpload" id="fileToUpload">
    <input type="submit" value="Upload Image" name="submit">


    Name of Image: <input type="text" name="imageName"/>

Answer 1

您需要指定它：

/* this->> */ $uploads_dir = '/uploads';
foreach ($_FILES["pictures"]["error"] as $key => $error) {
    if ($error == UPLOAD_ERR_OK) {
        $tmp_name = $_FILES["pictures"]["tmp_name"][$key];
        $name = $_FILES["pictures"]["name"][$key];
        /* then do this--> */ move_uploaded_file($tmp_name, "$uploads_dir/$name");
    }

因此，您将路径存储为$ uploads_dir 然后，您需要将文件移动到$ uploads_dir ...

那么，你在数据库中存储的路径将是=＆＃34; $ uploads_dir / $ name＆＃34;

将图像路径存储到数据库中

1 个答案: