对于循环执行两次，但仅保存数据一次PHP AJAX

  <html>
 <head>
 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
 </head>
 <body onload="searchVideo();">
<script>
 var data_length =  2;
 for(var i=0; i<data_length; i++){
console.log(i);
 var pageToken = '';
  var numOfResult = 0;
 var maxResults = 200;
 function searchVideo(){
   var separator = ',';
    $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken=' + pageToken + '&playlistId=PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-&key=Apikey&callback=?',function(data){
    var l = data.items.length;
    pageToken = data.nextPageToken;
    numOfResult += l;
    var itemUrl = '';
      var videoids = [];
         for(var j = 0; j < l; j++) {
          if( j == 0) {
            separator = ',';
          }
          else {
            separator = ',';
          }
            var videoid = data.items[j].snippet.resourceId.videoId;
            var title = data.items[j].snippet.title;
           $.ajax({
                method: 'POST',
               url:    'add.php',
               data: { title: title, videoid: videoid }
                 })
              .done(function(data) {
                });
        }
        if( numOfResult < maxResults) {
          searchVideo();
        }
         });
       }
   }
  </script>
</body>
 </html>

add.php

<?php
 ini_set('max_execution_time', 300);
include 'config.php';
 $title = mysqli_real_escape_string($mysql,$_POST['title']);
 $videoid = mysqli_real_escape_string($mysql,$_POST['videoid']);
 $thumbnail_url = 'http://img.youtube.com/vi/'.$videoid.'/hqdefault.jpg';
 $sql = "INSERT INTO ytfb(name,video_name,thumbnail_url)    VALUES('$title','$videoid','$thumbnail_url')";
$create_post_query=mysqli_query($mysql,$sql);
     if(!$create_post_query)
    {
   die("Connection failed".mysqli_error($mysql));
     }?>

从上面的代码数据中仅保存一次。数据没有保存两次，但正在运行两次FOR循环以保存数据两次。因此，任何人都可以帮助我使用Ajax两次保存数据。只有保存数据后，数据就不会保存两次，因为我给出的长度2表示循环2，应保存两次

。

根据我的评论，您的循环执行一次，而不是执行2次。如果要执行2次，则j<=1

在这一行中的问题。我已经解决了问题

for(var j = 0; j <= 1; j++) {

i在本地上解决了问题。您的函数searchVideo未执行两次。仅首次致电onload功能。我上传了一些代码示例。你应该休息这个样本

示例代码。删除功能并编写没有功能的代码

  <html>
 <head>
 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
 </head>
 <body>
<script>
 var data_length =  2;
 for(var i=0; i<2; i++){
    alert(i);
console.log(i);
 var pageToken = '';
  var numOfResult = 0;
 var maxResults = 200;

            var videoid = "asdf";
            var title = "erwe";
           $.ajax({
                method: 'POST',
               url:    'test4.php',
               data: { title: title, videoid: videoid }
                 })
              .done(function(data) {
                });

       }
  </script>
</body>
 </html>

test4.php

<?php
echo "string";
?>

实现功能的其他方法

 <html>
 <head>
 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
 </head>
 <body>
<script>
 var data_length =  2;
 for(var i=0; i<2; i++){
    searchVideo();
    alert(i)
console.log(i);
 var pageToken = '';
  var numOfResult = 0;
 var maxResults = 200;
 function searchVideo(){
            var videoid = "asdf";
            var title = "erwe";
           $.ajax({
                method: 'POST',
               url:    'test4.php',
               data: { title: title, videoid: videoid }
                 })
              .done(function(data) {
                });
        }
       }
  </script>
</body>
 </html>

尝试此

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
var paylistIdArray=['PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-','PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-'];
var pageToken ='';
var data_length =  paylistIdArray.length;
var numOfResult = 0;
var maxResults = 200;
var i=0;
var paylistId=paylistIdArray[0];
function searchVideo() {
    var separator = ',';
    $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken='+pageToken+'&playlistId='+paylistId+'&key= xxxx',function(data){
        console.log(data)
        var l = data.items.length;
        pageToken = data.nextPageToken;
        numOfResult += l;
        var itemUrl = '';
        var videoids = [];
        for(var j = 0; j < l; j++) {
            if( j == 0) {
                separator = ',';
            } else {
                separator = ',';
            }
            var videoid = data.items[j].snippet.resourceId.videoId;
            var title = data.items[j].snippet.title;
           console.log("called");
            $.ajax({
               method: 'POST',
               url:    'add.php',
               data: { title: title, videoid: videoid }
           }).done(function(data) {
           });
        }
        if( numOfResult < maxResults) {
          searchVideo();
        }
        if( numOfResult == maxResults && i<data_length) {
            i++;
            pageToken = ''; 
            numOfResult = 0;
            maxResults = 200;
            paylistId=paylistIdArray[i];
            searchVideo();
        }
    });
}
searchVideo()
</script>

no我们只称呼搜索视频功能一次。（在页面加载）从身体标签中删除Onload功能请尝试此代码

<script>
function searchVideo() {
    var separator = ',';
    $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken=' + pageToken + '&playlistId=PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-&key=Apikey&callback=?',function(data){
        var l = data.items.length;
        pageToken = data.nextPageToken;
        numOfResult += l;
        var itemUrl = '';
        var videoids = [];
        for(var j = 0; j < l; j++) {
            if( j == 0) {
                separator = ',';
            } else {
                separator = ',';
            }
            var videoid = data.items[j].snippet.resourceId.videoId;
            var title = data.items[j].snippet.title;
            $.ajax({
                method: 'POST',
                url:    'add.php',
                data: { title: title, videoid: videoid }
            }).done(function(data) {
            });
        }
        if( numOfResult < maxResults) {
          searchVideo();
        }
    });
}
var data_length =  2;
var pageToken = ''; 
var numOfResult = 0;
var maxResults = 200;
for(var i=0; i<data_length; i++){
    console.log(i);
    searchVideo();
}

  </script>

，但我不知道searchVideo函数中的以下条件

if( numOfResult < maxResults) {
          searchVideo();
        }

尝试此

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
var pageToken ='';
var data_length =  2;
var numOfResult = 0;
var maxResults = 200;
var i=1;
function searchVideo() {
    var separator = ',';
    $.getJSON('https://www.googleapis.com/youtube/v3/playlistItems?part=snippet&maxResults=50&pageToken='+pageToken+'&playlistId=PLrEnWoR732-BHrPp_Pm8_VleD68f9s14-&key= xxxx',function(data){
        console.log(data)
        var l = data.items.length;
        pageToken = data.nextPageToken;
        numOfResult += l;
        var itemUrl = '';
        var videoids = [];
        for(var j = 0; j < l; j++) {
            if( j == 0) {
                separator = ',';
            } else {
                separator = ',';
            }
            var videoid = data.items[j].snippet.resourceId.videoId;
            var title = data.items[j].snippet.title;
           console.log("called");
            $.ajax({
               method: 'POST',
               url:    'add.php',
               data: { title: title, videoid: videoid }
           }).done(function(data) {
           });
        }
        if( numOfResult < maxResults) {
          searchVideo();
        }
        if( numOfResult == maxResults && i<data_length) {
            i++;
          pageToken = ''; 
   numOfResult = 0;
   maxResults = 200;
          searchVideo();
        }
    });
}
searchVideo()
</script>

相关内容

最新更新

热门标签：