解析错误:语法错误,意想不到的'[' [英] Parse error: syntax error, unexpected '['
问题描述
我是那种新的使用PHP数组。现在我有一个小问题。我创建一个PHP文件,该文件会得到一些出击从YouTube的JSON的。现在,当我在我的本地运行我的应用程序运行得很好,但是当我试图到我校的主机上运行它,我得到一个错误。
这是我的code:
< PHP
//德数据库toeschrijven。 ($就在main.js !!!!)!
如果(使用isset($ _ POST ['zoekTerm'])){
$ zoekterm = $ _ POST ['zoekTerm'];//查询uitvoeren!
$添加= mysqli_query($康恩,/ *我的SQL查询* /);// JSON lijst inladen连接ALS查询德variabel $ zoekterm gebruiken。
$ tubeApiUrl = file_get_contents('http://gdata.youtube.com/feeds/api/videos?alt=json&q='.$zoekterm.'&orderby=published&max-results=3&v=2');
$ tube_json = json_de code($ tubeApiUrl,真正的); $ flickrApiUrl = file_get_contents('http://api.flickr.com/services/feeds/photos_public.gne?tags='.$zoekterm.'&format=json&nojsoncallback=1');
$ flickr_json = json_de code($ flickrApiUrl,真正的);// Compleet概述简单描述YOUTUBE,JSON,注释掉laten indien Niet的nodig。是handig嗡去JSON树TE volgen。
//回声< pre>中;
//的print_r($ tube_json);
//回声< / pre>中;
//每JSON入门EEN视频uitladen恩AAN EEN variabel林肯!
$ HIT1 = $ tube_json ['饲料'] ['进入'] [0] ['媒体集团$'] ['YT $ VideoID的'] ['$ T'];
$ HIT2 = $ tube_json ['饲料'] ['进入'] [1] ['媒体集团$'] ['YT $ VideoID的'] ['$ T'];
$ HIT3 = $ tube_json ['饲料'] ['进入'] [2] ['媒体集团$'] ['YT $ VideoID的'] ['$ T'];
$ hit4 = $ flickr_json ['项目'] [0] ['链接'];
$ hit5 = $ flickr_json ['项目'] [1] ['链接'];
$ hit6 = $ flickr_json ['项目'] [2] ['链接'];
//在EEN阵列stoppen视频的模具UIT JSON zijn gehaald。
$阵列= [$ HIT1,$ HIT2,$ HIT3,$ hit4,$ hit5,$ hit6]
$ tube_json_en code = json_en code($数组);
//德terug GE codeerde JSON printen NAAR谷歌浏览器控制台(网络)!
的print_r($ tube_json_en code);
和则错误:
解析错误:语法错误,意想不到的'['在第44行/home/stud/0861108/public.www/herkansing_imp3/back-end/add_tag.php
块引用>解决方案我不知道你用的是什么版本的PHP,使用方括号
[]
数组仅支持在PHP 5.4$阵列= [$ HIT1,$ HIT2,$ HIT3,$ hit4,$ hit5,$ hit6]
使用正常的方式,如果你是该版本在
$阵列=阵列($ HIT1,$ HIT2,$ HIT3,$ hit4,$ hit5,$ hit6);
I am sort of new using PHP arrays. Now I got a little problem. I create a PHP file that will get some hits out of the JSON from YouTube. Now when I run my application on my localhost it runs just fine, but when I'm trying to run it on my school's host, I get an error.
This is my code:
<?php //de database toeschrijven. ($.ajax in main.js!!!!) !! if(isset($_POST['zoekTerm'])){ $zoekterm = $_POST['zoekTerm']; //Query uitvoeren!!! $add = mysqli_query ($conn,"/*My SQL Query*/"); //JSON lijst inladen en als query de variabel $zoekterm gebruiken. $tubeApiUrl = file_get_contents('http://gdata.youtube.com/feeds/api/videos?alt=json&q='.$zoekterm.'&orderby=published&max-results=3&v=2'); $tube_json = json_decode($tubeApiUrl, true); $flickrApiUrl = file_get_contents('http://api.flickr.com/services/feeds/photos_public.gne?tags='.$zoekterm.'&format=json&nojsoncallback=1'); $flickr_json = json_decode($flickrApiUrl, true); //Compleet overzicht YOUTUBE-JSON, uncommented laten indien niet nodig. Is handig om de JSON tree te volgen. //echo "<pre>"; //print_r ($tube_json); //echo "</pre>"; //Per JSON entry een video uitladen en aan een variabel linken! $hit1 = $tube_json['feed']['entry'][0]['media$group']['yt$videoid']['$t']; $hit2 = $tube_json['feed']['entry'][1]['media$group']['yt$videoid']['$t']; $hit3 = $tube_json['feed']['entry'][2]['media$group']['yt$videoid']['$t']; $hit4 = $flickr_json['items'][0]['link']; $hit5 = $flickr_json['items'][1]['link']; $hit6 = $flickr_json['items'][2]['link']; //Video's die uit JSON zijn gehaald in een Array stoppen. $array = [$hit1, $hit2, $hit3, $hit4, $hit5, $hit6]; $tube_json_encode = json_encode($array); //De terug gecodeerde JSON printen naar Google Chrome Console (Network)!!! print_r ($tube_json_encode);
And then the error:
Parse error: syntax error, unexpected '[' in /home/stud/0861108/public.www/herkansing_imp3/back-end/add_tag.php on line 44
解决方案I wonder what version of php you are using, use square brackets
[]
for array is only supported in php 5.4$array = [$hit1, $hit2, $hit3, $hit4, $hit5, $hit6];
Use normal way if you are under that version
$array = array($hit1, $hit2, $hit3, $hit4, $hit5, $hit6);
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