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php - 使用ajax和PHP上传图像

转载 作者：行者123 更新时间：2023-11-29 11:35:22 24 4
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这是我的问题。我尝试上传图像，并借助此 URL 将图像存储在数据库中的相应文件夹和图像路径中。

http://www.lionblogger.com/how-to-upload-file-to-server-using-php-save-the-path-in-mysql/ 。

当我尝试这样做时，我会得到正确的答案。在我尝试使用 AJAX 时，相同的代码无法正常工作。我不知道我犯了什么错误。下面是我的代码。

HTML 代码

<div class="input-group form-group">
<label> Upload Your Photo </label>                            
<input type="file" name="upload_photo" id="upload_photo">
</div>
<div class="">
<input type="submit"  class="btn btn-success btn-lg " name="upload_files" id="upload_files" value="UPLOAD" >
</div>

AJAX代码

$("#upload_files").click(function(event){
        event.preventDefault();
        var upload_photo1 = $('#upload_photo').val();
        var photo= upload_photo1.split('\\').pop().split('/').pop();
        var datas="photo="+photo;
    alert(datas);
     if(photo==''){
      sweetAlert({
        title: "WARNING!!!",
        text: "Please Upload All Corresponding Documents And Try Again !!!!",
        type: "warning"
      });
     } else {
    $.ajax({
        type: "POST",
        url: 'php/upload_files.php',
        data:datas
        }).done(function( data ) {
            alert(data);
         });
    }
});

还有我的 PHP 文件上传文件.php

<?php
    $fileExistsFlag = 0; 
    $fileName = $_POST['photo'];
    var_dump($fileName);
    $link = mysqli_connect("localhost","root","","spark") or die("Error ".mysqli_error($link));

    $query = "SELECT filename FROM filedetails WHERE filename='$fileName'"; 
    $result = $link->query($query) or die("Error : ".mysqli_error($link));
    while($row = mysqli_fetch_array($result)) {
        if($row['filename'] == $fileName) {
            $fileExistsFlag = 1;
        }       
    }

    if($fileExistsFlag == 0)
    { 
        $target = "files/";     
        $fileTarget = $target.$fileName;    
        $tempFileName = $_FILES["fileName"]["tmp_name"];
        $fileDescription = $_POST['Description'];   
        $result = move_uploaded_file($tempFileName,$fileTarget);
        $ext = end(explode('.', $fileName));
        if ($_FILES["fileName"]["size"] > 2097152)
         {
         echo "Sorry, your file is too large.";
         $uploadOk = 0;
         }
      else if($ext != "jpg" && $ext != "png" && $ext != "jpeg"&& $ext != "gif" ) 
      {
         echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
         $uploadOk = 0;
      }
    else
    {
        if($result) { 
            echo "Your file <html><b><i>".$fileName."</i></b></html> has been successfully uploaded";       
            $query = "INSERT INTO filedetails(filepath,filename,description) VALUES ('$fileTarget','$fileName','$fileDescription')";
            $link->query($query) or die("Error : ".mysqli_error($link));            
        }
        else {          
            echo "Sorry !!! There was an error in uploading your file";         
        }
    }   
        mysqli_close($link);
    }

    else {
        echo "File <html><b><i>".$fileName."</i></b></html> already exists in your folder. Please rename the file and try again.";
        mysqli_close($link);
    }   
?>

我得到的错误是 enter image description here

请忍受我的疑问.. PHP 初学者。请帮助我解决这个问题。

最佳答案

您的 js 和 php 代码都存在严重问题。您根本没有上传文件。您刚刚上传了文件名。没有描述作为输入，但在您的 php 代码中，您有一个未处理的描述变量。您处理的文件名称错误。
尝试使用 FormData 上传文件。首先，使用带有 id 的 form 标签

<form id='myform'>
<div class="input-group form-group">
<label> Upload Your Photo </label>                            
<input type="file" name="upload_photo" id="upload_photo">
</div>
<div class="">
<input type="submit"  class="btn btn-success btn-lg " name="upload_files" id="upload_files" value="UPLOAD" >
</div>
</form>

在js中，

    formData = new FormData($("#myForm")[0]);
    formData.append("photo", photo);//your photo name

并在AJAX请求中写入

 data: formData,

在你的 php 文件中

<?php
 $fileExistsFlag = 0; 
 $fileDescription ='No idea where this came from';
 $fileName = $_POST['photo'];
 $link = mysqli_connect("localhost","root","","spark") or die("Error ".mysqli_error($link));

 $query = "SELECT filename FROM filedetails WHERE filename='$fileName'"; 
 $result = $link->query($query) or die("Error : ".mysqli_error($link));
 while($row = mysqli_fetch_array($result)) {
     if($row['filename'] == $fileName) {
         $fileExistsFlag = 1;
     }       
 }

 if($fileExistsFlag == 0)
 { 
     $target = "files/";     
     $fileTarget = $target.$fileName;    
     $tempFileName = $_FILES["upload_photo"]["tmp_name"];
     $result = move_uploaded_file($tempFileName,$fileTarget);
     $ext = end(explode('.', $fileName));
     if ($_FILES["upload_photo"]["size"] > 2097152)
      {
      echo "Sorry, your file is too large.";
      $uploadOk = 0;
      }
   else if($ext != "jpg" && $ext != "png" && $ext != "jpeg"&& $ext != "gif" ) 
   {
      echo "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";
      $uploadOk = 0;
   }
 else
{
    if($result) { 
        echo "Your file <html><b><i>".$fileName."</i></b></html> has been successfully uploaded";       
        $query = "INSERT INTO filedetails(filepath,filename,description) VALUES ('$fileTarget','$fileName','$fileDescription')";
        $link->query($query) or die("Error : ".mysqli_error($link));            
    }
    else {          
        echo "Sorry !!! There was an error in uploading your file";         
    }
}   
    mysqli_close($link);
}

else {
    echo "File <html><b><i>".$fileName."</i></b></html> already exists in your folder. Please rename the file and try again.";
    mysqli_close($link);
}   
?>

关于php - 使用ajax和PHP上传图像，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36693432/

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