Warning: file_get_contents(/data/phpspider/zhask/data//catemap/1/php/274.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
将Javascript数组转换为PHP数组_Javascript_Php_Arrays - Fatal编程技术网

将Javascript数组转换为PHP数组

将Javascript数组转换为PHP数组,javascript,php,arrays,Javascript,Php,Arrays,我创建PHP scraper是为了从Javascript数组中获取值,如何将Javascript数组转换为PHP数组 <?php // example scraped array $str = 'var str_array = ["aa", "bb"]'; $str = str_replace("var ", '$', $str); eval($str); print_r($str_array); 应以结尾 <?php $str = 'var str_array = ["aa",

我创建PHP scraper是为了从Javascript数组中获取值,如何将Javascript数组转换为PHP数组

<?php
// example scraped array
$str = 'var str_array = ["aa", "bb"]';
$str = str_replace("var ", '$', $str); 
eval($str);
print_r($str_array);
应以
结尾

<?php
$str = 'var str_array = ["aa", "bb"]';
$str = str_replace("var ", '$', $str); 
eval($str.";");
print_r($str_array);

但我不知道你为什么选择这个。

通过ajax将javascript数组作为JSON字符串发布,并在服务器端进行处理

Javascript

var str_array = ["aa", "bb"];
var request = $.ajax({
  url: "test.php",
  method: "POST",
  data: { myData : JSON.stringify(str_array) },
  dataType: "html"
});
request.done(function( msg ) {
  // ajax response
});
request.fail(function( jqXHR, textStatus ) {
  alert( "Request failed: " + textStatus );
});
PHP(test.PHP)


但是js是客户端,在使用
print\r
转换数组的not
print
后,可能会重复您试图执行的操作?更新
eval($str)进入
评估($str;“”)我需要js字符串在服务器中处理,而不是客户端。
var str_array = ["aa", "bb"];
var request = $.ajax({
  url: "test.php",
  method: "POST",
  data: { myData : JSON.stringify(str_array) },
  dataType: "html"
});
request.done(function( msg ) {
  // ajax response
});
request.fail(function( jqXHR, textStatus ) {
  alert( "Request failed: " + textStatus );
});
$json = $_POST['myData'];
$myDataArray = json_decode($json,true);
print_r($myDataArray);