javascript函数onsubmit不工作

时间:2013-10-22 10:30:01

标签: javascript php submit

我正在尝试this tutorial。我创建了一个index.php,其中包含一个包含以下代码的表单。问题是表单根本没有提交。警报没有被调用。什么都没发生。

<?php
require_once ('/soapclient/inc_connection.php');
 function insert()
 // if(isset($_POST['submit']))
 {
 //Describing the Leads object and printing the array
 $describe = $connection->describeSObjects(array('Lead'));
 print_r($describe);

 //Create New Lead
 $leadFirstName = "Ahmed";
 $leadLastName = "Dirie";
 $leadCompany = "Pick Yours Up";
 $leadEmail = "adirie@pickyoursup.com";

 //Creating the Lead Object
 $lead = new stdClass;
 $lead->type = 'Lead';
 $lead->fields = array(
      'FirstName' => $leadFirstName,
      'LastName' => $leadLastName,
      'Company' => $leadCompany,
      'Email' => $leadEmail
 );

 //Submitting the Lead to Salesforce
 $result = $connection->create(array($lead), 'Lead');
 }
?>

HTML:

 <h3>Sign up for an evaluation:</h3>

<span id="php_code"> </span>
 <form action="https://docs.google.com/..../formResponse"  method="POST" id="" target="_self" ><ol style="padding-left: 0">

        <div class="ss-form-question errorbox-good">
 <input type="button" name="submit" value="Contact Us" id="ss"  onclick="return postToSql();"/>
 ....

JS:

<script type="text/javascript" src="libs/jquery-1.7.1.js" ></script>
         <script type="text/javascript">

 function postToSql()
 {
  alert('1');

 alert("<?php insert(); ?>");
 //return false;
  }
 </script>

1 个答案:

答案 0 :(得分:0)

可能是PHP函数“insert()”的输出值导致javascript错误。所以你的javascript函数似乎未定义。尝试将insert()函数更改为此,您将看到差异。

function insert()
{
    echo "something to alert";
}

然后,

function insert()
{
    echo 'something " to alert';
}

要解决此问题,请确保您的函数的输出值不会因为Quotation(“)和Apostrophe(')Marks而导致错误。

如果你需要像这样使用Javascript和PHP,我建议使用Ajax。