Question

我在joomla 2.5模板文件中设置了这个PHP变量

$datestart = date('Y-m-d g:i', strtotime("+1 week"));
$dateend = date('Y-m-d g:i', strtotime("+30 days"));

当我回应这个

<?php echo json_encode($datestart); ?>

在php页面中它可以工作。我试图将此变量包含在RS Form组件的jQuery函数中

$formLayout .= "\n".'<script type="text/javascript" src="'.JURI::root(true).'/components/com_rsform/assets/calbs.js"></script>'."\n";

在.js文件中我有

jQuery(function() {
jQuery("#datetimepicker").datetimepicker({
format: "dd MM yyyy - HH:ii P",
showMeridian: true,
autoclose: true,
startDate: <?php echo json_encode($datestart); ?>,
endDate: <?php echo json_encode($dateend); ?>,
minuteStep: 15,
pickerPosition: "bottom-left"

});
 });

但它不起作用并显示php回显线的语法错误。我该如何解决这个问题？

Answer 1

您必须首先将php值分配给.php文件中的js变量

<script>
START_Date='<?php echo json_encode($datestart); ?>';
END_Date='<?php echo json_encode($dateend); ?>';

</script>

现在在你的js文件中

jQuery(function() {
    jQuery("#datetimepicker").datetimepicker({
        format: "dd MM yyyy - HH:ii P",
        showMeridian: true,
        autoclose: true,
        startDate: START_Date,
        endDate: END_Date,
        minuteStep: 15,
        pickerPosition: "bottom-left"

    });
 });

或者在你的php文件中包含你的js代码然后你可以在你的js中直接使用php变量：

jQuery(function() {
    jQuery("#datetimepicker").datetimepicker({
        format: "dd MM yyyy - HH:ii P",
        showMeridian: true,
        autoclose: true,
        startDate: '<?php echo json_encode($datestart); ?>',
        endDate: '<?php echo json_encode($dateend); ?>',
        minuteStep: 15,
        pickerPosition: "bottom-left"

    });
});

Answer 2

嗨你不能在js文件中使用php代码。将值存储在javascript变量中，然后在jquery函数中使用此变量。

Answer 3

您只需将此值存储到隐藏字段中，然后通过其id值将其输入.js文件。

Answer 4

试试这段代码

<input type="hidden" id="startdate" value="<?php echo json_encode($datestart); ?>"/>
<input type="hidden" id="enddate" value="<?php echo json_encode($dateend); ?>"/>



jQuery(function() {
var start = jQuery("#startdate").val();
var end= jQuery("#enddate").val();
 jQuery("#datetimepicker").datetimepicker({
  format: "dd MM yyyy - HH:ii P",
  showMeridian: true,
  autoclose: true,
  startDate: start,
  endDate: end,
  minuteStep: 15,
  pickerPosition: "bottom-left"
});
});

Answer 5

you should do it this way:
<?php

$datestart = date('Y-m-d g:i', strtotime("+1 week"));
$dateend = date('Y-m-d g:i', strtotime("+30 days"));

?>

 <input id="datestart" type="hidden" name="datestart" value="<?php echo $datestart; ?>">
  <input id="dateend" type="hidden" name="dateend" value="<?php echo $dateend; ?>">

  <script>
      var datestart = document.getElementById('datestart').value;
       var dateend = document.getElementById('dateend').value;

      jQuery(function() {
jQuery("#datetimepicker").datetimepicker({
format: "dd MM yyyy - HH:ii P",
showMeridian: true,
autoclose: true,
startDate: datestart,
endDate: dateend,
minuteStep: 15,
pickerPosition: "bottom-left"

});
 });
  </script>

在jQuery函数中回显php变量

5 个答案: