嘗試更新從while循環

Question

我的主要問題，我運行到生成的結果基本上是這樣的：

我有一個while循環，從查詢結果產生。通過生成的結果，我希望能夠更新原始查詢所來自的表格。

該查詢產生了預期的結果，但是當我單擊「刪除」按鈕時表格未被更新。我也試圖找到更新查詢執行後的結果要更新的解決方案......

<?php 
    $sql = "SELECT * FROM vehicles WHERE sold='n' ORDER BY year DESC"; 

    $query = mysql_query($sql); 


    while ($row = mysql_fetch_array($query)) { 

    echo 
    "  
     <tr> 
      <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['year'],"</td> 
      <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['make'],"</td> 
      <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['model'],"</td> 
      <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'><input type='submit' name='remove' value='REMOVE' style='background-color:#C33;color:white;padding:10px;border-radius:5px;width:70px'/></td> 
     </tr>"; 

    if(isset($_POST['remove'])){ 
     $removeSql = "UPDATE `table`.`vehicles` SET `display`='0' WHERE `vin`='{$row['vin']}'"; 
     mysql_query($removeSql) or die('check that code dummy'); 
    } 

    } 
    mysql_close($connection); 
?> 

Answer 1

這是一個提交按鈕，沒有表單標籤將無法正常工作。你不能這樣做。

您可以在單獨的頁面上寫remove code和轉換是submit button到normal button和使用ajax上button和call該頁面的click通過vin id。

或者，如果你不知道ajax和想要做的是頁面本身，然後做這種方式：

<?php 
$sql = "SELECT * FROM vehicles WHERE sold='n' ORDER BY year DESC"; 

$query = mysql_query($sql); 


while ($row = mysql_fetch_array($query)) { 

echo 
"  
    <tr> 
     <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['year'],"</td> 
     <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['make'],"</td> 
     <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'>",$row['model'],"</td> 
     <td style='border-bottom-style:dotted;padding-top:10px;padding-bottom:10px;font-size:.9em'> 
     <form action="" method="POST"> 
      <input type="hidden" name="vin_id" value="<?php echo $row['vin']; ?>"> 
      <input type='submit' name='remove' value='REMOVE' style='background-color:#C33;color:white;padding:10px;border-radius:5px;width:70px'/> 
     </form></td> 
    </tr>"; 

} 

if(isset($_POST['remove'])){ 
    $removeSql = "UPDATE `table`.`vehicles` SET `display`='0' WHERE `vin`='".$_POST['vin_id']."'"; 
    mysql_query($removeSql) or die('check that code dummy'); 
} 

mysql_close($connection); 
?>