ajax上傳多圖到php服務(wù)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="文件不超過200kb,大小最佳為60*60"> 
 </div> 
 <div class="img-box"> 
 <input type="file" name="photo2" id="" title="文件不超過200kb,大小最佳為60*60"> 
 </div> 
 </div> 
 <input type="button" class="bookBtn btnBlue" id="publishBook" value="發(fā)布圖書" 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="' + json['photo1'] + '" height="100" />'; 
 result += '<br/><img src="' + json['photo2'] + '" height="100" />'; 
 result += '<br/>' + json['photo1']; 
 result += '<br/>' + json['photo2']; 
 $('#result').html(result); 
 } 
 }); 
 return false; 
} 

<?php 
 require('conn.php'); 
 $nameTag = time(); 
 $filename1 = $nameTag . '0' . substr($_FILES['photo1']['name'], strrpos($_FILES['photo1']['name'],'.')); 
 $filename2 = $nameTag . '1' . substr($_FILES['photo2']['name'], strrpos($_FILES['photo2']['name'],'.')); 
 $response = array(); 
 $path1 = "img/" . $filename1; <span style="color:#ff0000;">//注意要在目錄下新建一個(gè)名為img的文件夾用來存放圖片 
 $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); 
?> 

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

<?php 
 require('conn.php'); 
 $image = mysql_real_escape_string(file_get_contents($_FILES['photo']['tmp_name'])); 
 $sqlstr = "insert into user(photo) values('".$image."')"; 
 @mysql_query($sqlstr) or die(mysql_error()); 
 exit(); 
?>