Question

我试图在Jquery中使用append 像这样

$('.div_addfield').append('<?php echo $this->Form->input("dob",array("separator"=>"","type"=>"date","label"=>"Date of Birth"));?>');

它没有在Firebug控制台上工作，它会给出以下错误

SyntaxError：unterminated string literal

但它显示数组如下

$('.div_addfield').append('<select id="ListOfStudentListOfHouseId" name="data[ListOfStudent]['+field_count+'][list_of_house_id]"><option value="1">Aitchison</option><option value="2">Nalagarh</option><option value="4">Ranjit</option><option value="5">Patiala</option></select>');
$('.div_addfield').append('<div class="input date"><label for="dobMonth">Date of Birth</label><select name="data[dob][month]" id="dobMonth">
<option value="01">January</option>
<option value="02">February</option>
<option value="03">March</option>
<option value="04">April</option>
<option value="05">May</option>
<option value="06">June</option>
<option value="07">July</option>
<option value="08" selected="selected">August</option>
<option value="09">September</option>
<option value="10">October</option>
<option value="11">November</option>
<option value="12">December</option>
</select><select name="data[dob][day]" id="dobDay">
<option value="01">1</option>
<option value="02">2</option>
<option value="03">3</option>
<option value="04">4</option>
<option value="05">5</option>
<option value="06">6</option>
<option value="07">7</option>
<option value="08">8</option>
<option value="09">9</option>
<option value="10">10</option>
<option value="11">11</option>
<option value="12">12</option>
<option value="13">13</option>
<option value="14">14</option>
<option value="15">15</option>
<option value="16">16</option>
<option value="17">17</option>
<option value="18">18</option>
<option value="19">19</option>
<option value="20">20</option>
<option value="21">21</option>
<option value="22">22</option>
<option value="23">23</option>
<option value="24">24</option>
<option value="25">25</option>
<option value="26">26</option>
<option value="27" selected="selected">27</option>
<option value="28">28</option>
<option value="29">29</option>
<option value="30">30</option>
<option value="31">31</option>
</select><select name="data[dob][year]" id="dobYear">
<option value="2032">2032</option>
<option value="2031">2031</option>
<option value="2030">2030</option>
<option value="2029">2029</option>
<option value="2028">2028</option>
<option value="2027">2027</option>
<option value="2026">2026</option>
<option value="2025">2025</option>
<option value="2024">2024</option>
<option value="2023">2023</option>
<option value="2022">2022</option>
<option value="2021">2021</option>
<option value="2020">2020</option>
<option value="2019">2019</option>
<option value="2018">2018</option>
<option value="2017">2017</option>
<option value="2016">2016</option>
<option value="2015">2015</option>
<option value="2014">2014</option>
<option value="2013">2013</option>
<option value="2012" selected="selected">2012</option>
<option value="2011">2011</option>
<option value="2010">2010</option>
<option value="2009">2009</option>
<option value="2008">2008</option>
<option value="2007">2007</option>
<option value="2006">2006</option>
<option value="2005">2005</option>
<option value="2004">2004</option>
<option value="2003">2003</option>
<option value="2002">2002</option>
<option value="2001">2001</option>
<option value="2000">2000</option>
<option value="1999">1999</option>
<option value="1998">1998</option>
<option value="1997">1997</option>
<option value="1996">1996</option>
<option value="1995">1995</option>
<option value="1994">1994</option>
<option value="1993">1993</option>
<option value="1992">1992</option>
</select></div>');

任何人都知道如何解决此错误

谢谢

Answer 1

你不能在JavaScript中使用跨越多行的字符串，尽管你可以在行的末尾添加\，字符串将在下一行继续。

Answer 2

尝试使用此

$('.div_addfield').append('<?php echo  str_replace(array("\r\n", "\r", "\n"), "",$this->Form->input("dob",array("separator"=>"","type"=>"date","label"=>"Date of Birth")));?>');

CakePHP和Jquery附加

2 个答案: