php如何实现post跳转

<?php
//定义post处理页面url
$actionUrl = './action.php' ;
 
//获取当前页面的url
$url = 'http://' . $_SERVER [ 'HTTP_HOST' ]. $_SERVER [ 'PHP_SELF' ]. '?' . $_SERVER [ 'QUERY_STRING' ];
 
//解析url
$parseInfo = parse_url ( $url );
 
//获取url中的参数部分
$queryString = $parseInfo [ 'query' ];
 
//将querycan参数解析到数组变量
$queryArr =[];
parse_str ( $queryString , $queryArr );
 
//循环拼接表单项
$formItemString = '' ;
foreach ( $queryArr as $key => $value ){
 $formItemString .= "<input name='{$key}' type='text' value='{$value}'/>" ;
}
 
//构造表单并跳转
$content =<<<EOF
<form style= 'display:none' name= 'submit_form' id= 'submit_form' action= '{$actionUrl}' method= 'post' >
{ $formItemString }
</form>
<script type= "text/javascript" >
document.submit_form.submit();
</script>
EOF;
 
exit ( $content );

<?php
//打印post接收到的数据
echo print_r( $_POST ,true);