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開關輸入鏈接我有一個cHTML等::鏈接是這樣的:的Yii了CHtml ::裏面
echo CHtml::link('DESACTIVAR',array ('/ZfIncidencias/estado', 'id'=>$data->incidencia_id),array("class"=>"desactivar"));
此執行動作,但我需要的是這樣的:
$id_on = $data->incidencia_id;
$content_on = '<div class="onoffswitch">
<input type="checkbox" name="onoffswitch" class="onoffswitch-checkbox" id="'.$id_on.'" checked />
<label class="onoffswitch-label" for="'.$id_on.'">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div> ';
echo CHtml::link($content_on,array('/ZfIncidencias/estado', 'id'=>$data->incidencia_id));
當我單擊該操作未執行。
有辦法迫使它執行?
SOLUTION:
$id_off = $data->incidencia_id;?>
<div class="onoffswitch">
<input type="checkbox" name="onoffswitch" class="onoffswitch-checkbox" id="<?php echo $id_off;?>" unchecked onClick="javascript:location.href = '<?php echo Yii::app()->baseUrl.'/ZfIncidencias/estado/'.$data->incidencia_id;?>'"; />
<label class="onoffswitch-label" for="<?php echo $id_off;?>">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div>
是有一個原因的複選框必須在鏈接中? – Goodnickoff
因爲是一個切換按鈕,並且它在點擊時發生變化 –
但是爲什麼在這個按鈕中是複選框? – Goodnickoff