Question

我有一個表單提交2個值到我的數據庫，當表單加載時我檢查數據庫中是否已經存在值，我一直在嘗試使用這些值並將它們設置爲我的默認值下拉框。任何幫助，將不勝感激。感謝

<form method="post" action="add_res.php?id=<? echo $id;?>"> 
    <fieldset> 
    <label>From</label> 
    <select name='one' id='one'> 
    <? 
     for($hours=0; $hours<24; $hours++) // the interval for hours is '1' 
      for($mins=0; $mins<60; $mins+=30) // the interval for mins is '30' 
       echo '<option>'.str_pad($hours,2,'0',STR_PAD_LEFT).':'.str_pad($mins,2,'0',STR_PAD_LEFT).'</option>'; 
    ?> 
    </select><br><br> 
    <select name='two' id='two'> 
    <? 
     for($hours=0; $hours<24; $hours++) 
      for($mins=0; $mins<60; $mins+=30) 
       echo '<option>'.str_pad($hours,2,'0',STR_PAD_LEFT).':'.str_pad($mins,2,'0',STR_PAD_LEFT).'</option>'; 
    ?> 
    </select> 
    </fieldset> 
    <input type="submit" value="Restrict" class="btn_searchp"></a> 
</form> 

Answer 1

你的價值結合起來，然後與數據庫中的值進行比較

<select name="one" id="one"> 
<?php for loop 
$timeToDisplay = sprintf('%s:%s',str_pad($hours,2,'0',STR_PAD_LEFT), 
           str_pad($mins,2,'0',STR_PAD_LEFT)); 
$selected = if($row['timeRow'] == $timeToDisplay) ? 'selected="selected"' : ''; 
// let $row['timeRow'] is the value retrieve from database 
?> 
    <option value="<?=php echo $timeToDisplay;?>" <?php echo $selected;?> > 
    <?php echo $timeToDisplay?> 
    </option> 
</select> 

參考文獻：

sprintf

ternary operator