Question

当我添加表单验证jquery时，我有一个停止按预期运行的表单。

<form class="myform" action="booking.php?token=<?php echo time(); ?>" method="get" name="f<?php echo ( $row['vehicle_id'] );?>" class="request_submit">
    <input type="hidden" name="v_id" value="<?php echo ( $row['vehicle_id'] );?>" />
    <input type="hidden" name="p_id" value="<?php echo $params['quote_location']['pickup']['location_id']; ?>" />
    <input type="hidden" name="p_val" value="<?php echo $params['quote_location']['pickup']['location_name']; ?>" />
    <input type="hidden" name="d_id" value="<?php echo $params['quote_location']['dropp']['location_id']; ?>" />
    <input type="hidden" name="d_val" value="<?php echo $params['quote_location']['dropp']['location_name']; ?>" />                     
<?php
    if (  $row['vehicle_price_mode'] == 'fixed' ):
        $v_price = $row['vehicle_price'] ;
    else:
        $v_price = ( $params['quote_price'] * $row['vehicle_price'] ) / 100 ;
    endif;

?>
<div style="padding:15px;">
    <table style="width:200px;border:0px solid black">
    <tr>
        <td align="right">
            <div class="quoteradio">
                <label><input name="q_price" type="radio" checked="checked" value="oneway|<?php echo swd_number_format(  ( $params['quote_price'] + $v_price )); ?>"> One Way: &pound;<?php echo (  swd_number_format(  $params['quote_price'] + $v_price  ) ); ?></label>
            </div>  
        </td>
    </tr>
    <tr>
        <td align="right">
            <div class="quoteradio">
                <label><input name="q_price" type="radio" value="twoway|<?php echo swd_number_format(  ( $params['quote_price'] + $v_price ) * 2 ); ?>"> Two Way: &pound;<?php echo swd_number_format(  ( $params['quote_price'] + $v_price ) * 2 ) ; ?></label>
            </div>
        </td>
    </tr>
    <tr>
        <td align="right">                      <button  type="submit" class="btn btn-primary">RESERVE </button></td>
    </tr>
    </table> 
</div>

链接到提交操作

$(".myform").submit(function(e){
    var targeted_popup_class = "popup-1";
    $('[data-popup="' + targeted_popup_class + '"]').fadeIn(350);

    e.preventDefault();
    var vehicle_id = $(this).parent().find('input[type="hidden"][name="v_id"]').val();
    $('input[name="vehicle_id"]').val(vehicle_id);
    var pickup_id = $(this).parent().find('input[type="hidden"][name="p_id"]').val();
    $('input[name="pickup_id"]').val(pickup_id);    
    var pickup_name = $(this).parent().find('input[type="hidden"][name="p_val"]').val();
    $('input[name="pickup_name"]').val(pickup_name);
    var dest_id = $(this).parent().find('input[type="hidden"][name="d_id"]').val();
    $('input[name="dest_id"]').val(dest_id);
    var dest_name = $(this).parent().find('input[type="hidden"][name="d_val"]').val();
    $('input[name="dest_name"]').val(dest_name);
    var quote_price = $(this).parent().find('input[type="radio"][name="q_price"]:checked').val();
    $('input[name="quote_price"]').val(quote_price);
});

但是当我尝试在不同表单的同一页面中包含表单验证时我的初始提交jquery停止按预期运行

只是在我之前的jquery上面添加这个jquery税就会停止它按预期运行

      $("form[name='registration']").validate({

    }
  });

。请告诉我为什么以及如何解决这个问题。

由于

添加表单验证jquery会导致以前的jquery提交函数无效

0 个答案: