在php文件中验证开始日期和结束日期时出错

Posted 2021-04-07

当我们声明java脚本验证没有应用而没有触发错误时，ws与输入类型相同，我已经在php文件中声明了代码。

我使用过Xammp server，运行mysql 5，PHP 7.6.2和Apache 2。

<script  type="text/javascript">
    //Validation for Stratdate & Enddate for New Ticket creation form                       

    $("#tedate").change(function () {

        var objFromDate = document.getElementById("tsdate").value; 
        var objToDate = document.getElementById("tedate").value;

        var FromDate = new Date(objFromDate);
        var ToDate = new Date(objToDate);

        if(FromDate > ToDate )
        {
            alert("Due Date Should Be Greater Than Start Date");
            document.getElementById("tedate").value = "";
            return false; 
        }

    });
</script>

开始日期

<input type='date' id="tsdate" class="form-control col-md-6"  placeholder="mm-dd-yyyy" name="startdate">

截止日期

<input type='date' id="tedate" class="form-control col-md-6" placeholder="Enter Due Date" name="enddate" required />

我想验证到期日期大于开始日期

答案

在php中使用strtotime函数

if(strtotime($date1) < strtotime($date2)){
      /* Code */
}

另一答案

使用getTime（）函数将时间转换为纪元格式。有关详细信息，请访问https://www.epochconverter.com/programming/#javascript

您可以执行以下操作来检查前端。

使用Javascript

  $(".data-controller").change(function () {
        var objFromDate = document.getElementById("tsdate").value;
        var objToDate = document.getElementById("tedate").value;

        var FromDate = (new Date(objFromDate).getTime()) / 1000;
        var ToDate = (new Date(objToDate).getTime()) / 1000;

        if (FromDate > ToDate) {
            alert("Due Date Should Be Greater Than Start Date");
        }

    });

html

<input type='date' id="tsdate" class="form-control col-md-6 data-controller" placeholder="mm-dd-yyyy"
               name="startdate">  // class data-controller added
<input type='date' id="tedate" class="form-control col-md-6 data-controller" placeholder="Enter Due Date"
               name="enddate"
               required/> // class data-controller added