Question

在php中，我想使用jquery.post方法传递一个php变量。

第一张图显示了发布变量的代码 first

第二张图显示了阅读发布的变量。second

php变量包含一些语句，但输出将打印变量名称，如＆＃34; $ rxss＆＃34;，＆＃34; $ sxss＆＃34;。

post方法似乎是错误地写了数据部分，所以请让我知道如何编写它。

添加

我已经测试了

echo '<script type="text/javascript">
$(document).ready(function() {
$.post("scanner/getStatus.php", {testId:' . "$testId" . ',chkCnt:' . "$chkCount" . ',rxss:' . "$rxss" . ',sxss:' . "$sxss" . ',sqli:' . "$sqli" . ',basqli:' . "$basqli" . ',autoc:' . "$autoc" . ',idor:' . "$idor" . ',dirlist:' . "$dirlist" . ',bannerdis:' . "$bannerdis" . ',sslcert:' . "$sslcert" . ',unredir:' . "$unredir" . ',clamav:' . "$clamav" . '}, function(data){$("#status").html(data)});
  var refreshId = setInterval(function() {
$.post("scanner/getStatus.php", {testId:' . "$testId" . ',chkCnt:' . "$chkCount" . ',rxss:' . "$rxss" . ',sxss:' . "$sxss" . ',sqli:' . "$sqli" . ',basqli:' . "$basqli" . ',autoc:' . "$autoc" . ',idor:' . "$idor" . ',dirlist:' . "$dirlist" . ',bannerdis:' . "$bannerdis" . ',sslcert:' . "$sslcert" . ',unredir:' . "$unredir" . ',clamav:' . "$clamav" . '}, function(data){$("#status").html(data)});
}, 500); ......

-

echo '<script type="text/javascript">
$(document).ready(function() {
$.post("scanner/getStatus.php", {testId:' . "$testId" . ',chkCnt:' . "$chkCount" . ',rxss:"" + $rxss + "",sxss:"" + $sxss + "",sqli:""+ $sqli + "",basqli:"" + $basqli + "",autoc:"" + $autoc + "",idor:"" + $idor + "",dirlist:"" + $dirlist + "",bannerdis:"" + $bannerdis + "",sslcert:"" + $sslcert + "",unredir:"" + $unredir + "",clamav:"" + $clamav + ""}, function(data){$("#status").html(data)});
  var refreshId = setInterval(function() {
$.post("scanner/getStatus.php", {testId:' . "$testId" . ',chkCnt:' . "$chkCount" . ',rxss:"" + $rxss + "",sxss:"" + $sxss + "",sqli:""+ $sqli + "",basqli:"" + $basqli + "",autoc:"" + $autoc + "",idor:"" + $idor + "",dirlist:"" + $dirlist + "",bannerdis:"" + $bannerdis + "",sslcert:"" + $sslcert + "",unredir:"" + $unredir + "",clamav:"" + $clamav + ""}, function(data){$("#status").html(data)});
}, 500); ......

但是当这些测试没有显示getStatus.php和apache log时出现错误

Answer 1

你可以这样$("$yourvariable") 另请查看此How to call PHP function from string stored in a Variable

在javascript中发布到php变量

1 个答案: