douxiong3234 2014-12-01 11:10
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插入数据库

I have a litte problem. I have this code, but this insert in to database only last input value. My function doing a new input, if u click + then coming again new input, but I want all these inputs values in to database, but this code inserting only last input value in to database. Example: I inserting 53424345 and then 56546546, and only 56546546 goes in to database. But I want in to database 53424345, 56546546. Is that possible ? And how to do something like this ?

<table id="mytable">
        <tr id="number0" style="display:none;">
            <td><?php echo $this->Form->button('&nbsp;-&nbsp;',array('type'=>'button','title'=>'Click Here to remove this number')); ?></td>
            <td><?php echo $this->Form->input('lisanumbrid'); ?></td>
        </tr>
        <tr id="trAdd"><td> <?php echo $this->Form->button('+',array('type'=>'button','title'=>'Click Here to add another number','onclick'=>'addNumber()')); ?> </td><td></td><td></td><td></td><td></td></tr>
    </table>



    <?php echo $this->Html->script(array('jquery-2.1.1.min.js'));?>
<script type='text/javascript'>
    var lastRow=0;

    function addNumber() {
        lastRow++;
        $("#mytable tbody>tr#number0").clone(true).attr('id','lisanumbrid'+lastRow).removeAttr('style').insertBefore("#mytable tbody>tr#trAdd");
        $("#lisanumbrid"+lastRow+" button").attr('onclick','removeNumber('+lastRow+')');
        $("#lisanumbrid"+lastRow+" input:first").attr('numbrid','data[Lisanumbrid]['+lastRow+'][lisanumbrid]').attr('id','numbridlisaNumber'+lastRow);
    }

    function removeNumber(x) {
        $("#lisanumbrid"+x).remove();
    }
</script>

Thanks, I accept all hints and solutions.

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1条回答 默认 最新

  • doumin1897 2014-12-01 11:24
    关注

    You need to do 2 changes, considering you are saving data into Model table:

    1) Modify your input as

    <?php echo $this->Form->input('lisanumbrid',array('name'=>'data[Model][0][lisanumbrid]')); ?>
    

    2) Modify your addNumber() as

    function addNumber() {
            lastRow++;
            $("#mytable tbody>tr#number0").clone(true).attr('id','lisanumbrid'+lastRow).removeAttr('style').insertBefore("#mytable tbody>tr#trAdd");
            $("#lisanumbrid"+lastRow+" button").attr('onclick','removeNumber('+lastRow+')');
            $("#lisanumbrid"+lastRow+" input:first").attr('numbrid','data[Lisanumbrid]['+lastRow+'][lisanumbrid]').attr({'id':'numbridlisaNumber'+lastRow,'name': 'data[Model]['+ lastRow  +'][lisanumbrid]'});
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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