ajax上传多图到php服务器的方法

<form id="form1"> 
<span style="white-space:pre;"> </span><div class="bookImg"> 
    <div class="img-box"> 
      <input type="file" name="photo1" id="" title="文件不超过200kb,大小最佳为60*60"> 
    </div> 
    <div class="img-box"> 
      <input type="file" name="photo2" id="" title="文件不超过200kb,大小最佳为60*60"> 
    </div>               
  </div> 
  <input type="button" class="bookBtn btnBlue" id="publishBook" value="发布图书" onclick="fsubmit()"/> 
</form> 

function fsubmit() { 
  var form1=document.getElementById("form1"); 
    var fd =new FormData(form1); 
    $.ajax({ 
       url: "photo.php", 
       type: "POST", 
       data: fd, 
       processData: false, 
       contentType: false, 
       success: function(response,status,xhr){ 
        console.log(xhr); 
        var json=$.parseJSON(response); 
        var result = ''; 
         result += '<br/><img src="/UploadFiles/2021-04-02/' + json['photo1'] + '">

php代码：photo.php


<"img/" . $filename1; <span style="color:#ff0000;">//注意要在目录下新建一个名为img的文件夹用来存放图片 
    $path2 = "img/" . $filename2; 
    if(move_uploaded_file($_FILES['photo1']['tmp_name'], $path1) && move_uploaded_file($_FILES['photo2']['tmp_name'], $path2) ){            
      $response['isSuccess'] = true;   
      $response['photo1'] = $path1;  
      $response['photo2'] = $path2;       
    }else{  
      $response['isSuccess'] = false;  
    }  
    echo json_encode($response); 
"htmlcode">

<form action="photo.php"> 
<span style="white-space:pre;"> </span><div class="pic"> 
    <input type="file" name="photo" id="" title="文件不超过200kb,大小最佳为60*60" onchange="imgPreview(this)">上传头像 
  </div> 
</form> 


php代码：photo.php


<"insert into user(photo) values('".$image."')";        
  @mysql_query($sqlstr) or die(mysql_error());   
  exit();        
?> 


这样就把图片转换成二进制并储存进数据库了。
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。