2017-06-12 139 views
0

我有一个表单,它可以在另一个站点上执行GET操作,并且希望在提交表单时使用?_ga=跟踪器ID。我有这个跟踪器clientId正常工作正常链接,但不知何故无法让它在下面的窗体上工作。以表格形式设置Google Analytics(分析)跟踪器clientId

<form action="" method="get" name="iform" id="iform" target="_blank"> 
    <div class="form-items"> 
     <div class="form-item fas-feedback date date-1"> 
      <label for="DateIn">From</label> 
      <input type="text" class="check-date-from" id="check-in" name="DateIn" readonly placeholder="DD-MM-YYYY" /> 
     </div> 
     <div class="form-item fas-feedback date date-2"> 
      <label for="DateOut">Till</label> 
      <input type="text" class="check-date-to" id="check-out" readonly name="DateOut" placeholder="DD-MM-YYYY" /> 
     </div> 
    </div> 
    <div class="form-action long-transition"> 
     <button>Submit</button> 
    </div> 
</form> 

和脚本:

var clientId; 
$(document).ready(function() { 
    ga(function(tracker) { 
     clientId = tracker.get('clientId'); 
    }); 

    $("#iform").submit(function(e) { 
     e.preventDefault(); 
     var uri = 'https://engines.hoteliers.com/en/3456789/whitelabelplus/getdata/arrival/'+$("#check-in").val()+'/departure/'+$("#check-out").val()+'/type/roomonly/?clientId='+clientId; 
     window.open(uri, '_blank'); 
     return false; 
    }) 
}); 

任何帮助将是巨大的!

+0

你说的不工作呢?是'clientId'错误?或空或什么? – Huangism

+0

clientId不是空的,URL是用正确的'?clientId = 2345678'创建的,它只是不会将cookie正确设置为'?_ga = 90876543',而常规链接可以正常工作 –

回答

0

尝试

$(document).ready(function() { 
    $("#iform").submit(function(e) { 
     e.preventDefault(); 
     var clientId = ga.getAll()[0].get('clientId');  
     var uri = 'https://engines.hoteliers.com/en/3456789/whitelabelplus/getdata/arrival/'+$("#check-in").val()+'/departure/'+$("#check-out").val()+'/type/roomonly/?clientId='+clientId; 
     window.open(uri, '_blank'); 
     return false; 
    }); 
}); 
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