2016-11-12 96 views

回答

5

我只是找到了!

function postToGoogle() { 
 
    var field1 = $("input[type='radio'][name='qs1']:checked").val(); 
 
    var field2 = $('#feed').val(); 
 

 
    $.ajax({ 
 
    url: "https://docs.google.com/forms/d/e/1FAIpQLSdjOTKRb7YiWi8OGPq6M6CRL0TpuAsUKacKp2XgruMbIp4wzg/formResponse", 
 
    data: { 
 
     "entry.924752166": field1, 
 
     "entry.997497831": field2 
 
    }, 
 
    type: "POST", 
 
    dataType: "xml", 
 
    statusCode: { 
 
     0: function() { 
 
     //Success message 
 
     }, 
 
     200: function() { 
 
     //Success Message 
 
     } 
 
    } 
 
    }); 
 
} 
 

 
/* $(document).ready(function() { 
 
    $('#form').submit(function() { 
 
    postToGoogle(); 
 
    return false; 
 
    }); 
 
    
 
});*/
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 

 
<strong>Test Google Form</strong> 
 
<form id="form" target="_self" onsubmit="" action="javascript: postToGoogle()"> 
 
    <fieldset> 
 
    <label>Question 1</label> 
 
    <input id="qs1_op_1" type="radio" value="Yes" name="qs1" /> 
 
    <input id="qs1_op_2" type="radio" value="No" name="qs1" /> 
 
    </fieldset> 
 

 
    <fieldset> 
 
    <label>Text</label> 
 
    <textarea id="feed" name="feed"></textarea> 
 
    </fieldset> 
 
    <div style="width: 100%; display: block; float: right;"> 
 
    <button id="send" type="submit"> 
 
     Send 
 
    </button> 
 
    </div> 
 
</form> 
 
<br /><br /> 
 
The <a href="https://docs.google.com/spreadsheets/d/1bVeLfK2gm6emaGRKHnMllpeb_P4HwwZoIfZB5MCcyZg/pubhtml">Result</a> takes few minutes to be shown, but it is sent to the google sheet instantaneously. 
 
<br /><br />

它的工作在CodePen

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