试图将PHP的值转换为javascript

trying to get the value of php into javascript

本文关键字：转换 javascript PHP      更新时间：2023-09-26

我试图获得$dateyearrange的值放入yearRange:比如我的$dateyearrange = 2001那么yearRange:'dateyearrange:+0'应该是yearRange:'2001:+0'但我似乎不能使它工作

编辑:问题在我的代码是在yearRange:('dateyearrange:+0')

如果我输入echo <?php $Dateyear ?>，代码将不会执行因为如果我使用yearRange:dateyearrange，在yearrange上的输出只有2001，这是代码工作得很好。所以问题在我的代码是在yearRange:我想要在yearrange中显示的是2001:+0

我很抱歉我的英语，我不能真正地向它解释我自己。

编辑2:(解决方案):

我使用yearRange:"<?php echo $DateYear ?>:+0"并删除var dateyearrange = "<?php $Dateyear ?>";我只是让我的代码更复杂了，我不知道为什么要这样做。谢谢你的回复，它帮助了我。

<?php
$DateYear = date("Y", strtotime($Datebirth)) 
;?>
var dateminimum = "<?php Echo($Datebirth); ?>";
var dateyearrange = "<?php $Dateyear ?>";
  $(function() {
    $( "#datepicker" ).datepicker({minDate:dateminimum, maxDate:'0', yearRange:('dateyearrange:+0'), dateFormat: "yy-mm-dd",changeMonth: true,
      changeYear: true});
  });

<?php
$DateYear = date("Y", strtotime($Datebirth)) 
;?>
var dateminimum = <?php echo $Datebirth; ?>;
var dateyearrange = <?php echo $Dateyear ?>;
  $(function() {
    $( "#datepicker" ).datepicker({minDate:dateminimum, maxDate:'0', yearRange:('dateyearrange:+0'), dateFormat: "yy-mm-dd",changeMonth: true,
      changeYear: true});
  });

改为:

<?php
$DateYear = date("Y", strtotime($Datebirth)) 
;?>
var dateminimum = "<?php echo $Datebirth; ?>";
var dateyearrange = "<?php echo $Dateyear; ?>";
  $(function() {
    $( "#datepicker" ).datepicker({minDate:dateminimum, maxDate:'0', yearRange:('dateyearrange:+0'), dateFormat: "yy-mm-dd",changeMonth: true,
      changeYear: true});
  });

您仍然需要echo或print或= 

var dateminimum = "<?php echo $Datebirth; ?>";
var dateyearrange = "<?php echo $Dateyear; ?>";
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