尝试更新从while循环

Question

我的主要问题，我运行到生成的结果基本上是这样的：

我有一个while循环，从查询结果产生。通过生成的结果，我希望能够更新原始查询所来自的表格。

该查询产生了预期的结果，但是当我单击“删除”按钮时表格未被更新。我也试图找到更新查询执行后的结果要更新的解决方案......

<?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

这是一个提交按钮，没有表单标签将无法正常工作。你不能这样做。

您可以在单独的页面上写remove code和转换是submit button到normal button和使用ajax上button和call该页面的click通过vin id。

或者，如果你不知道ajax和想要做的是页面本身，然后做这种方式：

<?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); 
?>