如何使用Java脚本验证php表单，该表单在我网站的每个页面上

Question

The problem i'm having is Im trying to get the contact.php to validate a contact form that is now in the bottom of every page in my site. 我遇到的问题是我正在尝试获取contact.php来验证现在位于我网站每个页面底部的联系表单。 It worked great with just validating one page and having a hard URL to take them back to. 只需验证一个页面并使用硬URL即可将其带回，因此效果很好。 the troubleing code is this. 麻烦的代码是这个。

/* In case of no valid input go back  */
if ($p1==""||$p2==""||$p3==""||$p5==""||$n0!=0||$f0<7||$NL==0||$p7=="") {
    header($_SERVER['HTTP_REFERER'].$report); 
}else{$report='noerror';}

Ive also tryed this: 我也尝试过这样：

/* In case of no valid input go back  */
if ($p1==""||$p2==""||$p3==""||$p5==""||$n0!=0||$f0<7||$NL==0||$p7=="") {
    header($_SERVER['PHP_SELF'].$report); 
}else{$report='noerror';}

This is the original code that worked for one page: 这是适用于一页的原始代码：

/* In case of no valid input go back  */
if ($p1==""||$p2==""||$p3==""||$p5==""||$n0!=0||$f0<7||$NL==0||$p7=="") {
   header('Location:estimate_page.php?'.$report);
}else{$report='noerror';}

I could just make a new "contact.php" page for every page that has the form in the footer, but thats stupid if i can just make this work. 我可以为每个在页脚中具有表单的页面创建一个新的“ contact.php”页面，但是如果我可以进行这项工作，那就太愚蠢了。 Any ideas 有任何想法吗

Answer 1

You are thinking in the right direction with javascript/jQuery -- it will allow you to catch the invalid fields before submitting / changing pages. 您正在使用javascript / jQuery朝着正确的方向进行思考-它将允许您在提交/更改页面之前捕获无效字段。 Much better user experience. 更好的用户体验。 Only submit when fields are validated. 仅在验证字段时提交。

Here is a link for how to do simple field validation testing with jQuery: 以下是有关如何使用jQuery执行简单字段验证测试的链接：

jsFiddle Demo jsFiddle演示

HTML: HTML：

One: <input type="text" id="f1"><br />
Two: <input type="text" id="f2"><br />
Three: <input type="text" id="f3"><br />
Four: <input type="text" id="f4"><br />
<br />
<input type="button" id="mybutt" value="Go" />

javascript/jQuery: 使用Javascript / jQuery的：

var chkFld, arrAll = {'One':'f1','Two':'f2','Three':'f3','Four':'f4'};

$('#mybutt').click(function() {
    var errMsg='', badFlds='', firstBad='';
    for(var key in arrAll){
        chkFld = '#'+arrAll[key];
        $(chkFld).removeClass('error');
        if ($(chkFld).val() ==''){
            $(chkFld).addClass('error');
            //alert('Please complete field ' + arrAll[key] );
            errMsg += '*' + key + '\n';
            if (firstBad=='') firstBad=chkFld;
        }
    }
    if (errMsg != '') {
        alert(errMsg);
        $(firstBad).focus();
    }
}); //END mybutt.click

Note that the above example uses jQuery, so you must reference the jQ libraries (usually inside the head tags, something like this:) 请注意，上面的示例使用jQuery，因此您必须引用jQ库（通常在head标签内，如下所示：）

<head>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
</head>

EDIT: 编辑：

Here is what the full HTML would look like: 完整的HTML如下所示：

<?php
    //Any initial PHP code goes here
?>

<html>
    <head>
        <script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>

        <script type="text/javascript">
            //javascript
        var chkFld, arrAll = {'One':'f1','Two':'f2','Three':'f3','Four':'f4'};
        $(document).ready(function() {

                $('#mybutt').click(function() {
                    var errMsg='', badFlds='', firstBad='';
                    for(var key in arrAll){
                        chkFld = '#'+arrAll[key];
                        $(chkFld).removeClass('error');
                        if ($(chkFld).val() ==''){
                            $(chkFld).addClass('error');
                            //alert('Please complete field ' + arrAll[key] );
                            errMsg += '*' + key + '\n';
                            if (firstBad=='') firstBad=chkFld;
                        }
                    }
                    if (errMsg != '') {
                        alert(errMsg);
                        $(firstBad).focus();
                        return false;
                    }

                    //If it gets to here, it passed validation
                        $('#yourForm').submit();
                }); //END mybutt.click()

            }); //END $(document).ready()

        </script>
    </head>
<body>

    <form id="yourForm">
        One: <input type="text" id="f1" /><br />
        Two: <input type="text" id="f2" /><br />
        Three: <input type="text" id="f3" /><br />
        Four: <input type="text" id="f4" /><br />
    </form>
    <input type="button" id="mybutt" value="Click Me">

</body>
</html>

Note that, using the above scenario, no further PHP code would execute until the form is submitted, using the manual $('#yourForm').submit(); 请注意，在上述情况下，使用手动$('#yourForm').submit();表单之前，不会再执行其他PHP代码$('#yourForm').submit(); statement. 声明。

如何使用Java脚本验证php表单，该表单在我网站的每个页面上

问题描述

1 个解决方案

解决方案1
0 2014-08-26 02:28:42

如何使用Java脚本验证php表单，该表单在我网站的每个页面上

问题描述

1 个解决方案

解决方案1 0 2014-08-26 02:28:42

解决方案1
0 2014-08-26 02:28:42