如何更新值在我的数据库与新值从<select>和& lt; input>标签


How to update values in my DB with new values from <select> and <input> tags?

通过这种方式,我从数据库中获得结果并"打印"它们。但我不知道我将如何更新这些结果,当我按下提交按钮!!我只是需要一个下一步的想法。提前谢谢你!!

下面是我的代码示例…

<?php // DATABASE QUERY
      $query="SELECT countdown_module, hometeam_position
                FROM jos_gm_nextmatch
               WHERE id = 1";
      $result=mysql_query($query);
      // DATABASE VARIABLES
      $countdown_module = mysql_result($result,$i,"countdown_module");
      $hometeam_position = mysql_result($result,$i,"hometeam_position"); ?>
<form action="***.php" method="post" name="form">
<input name="countdown_module" value="<?php echo $countdown_module ?>" type="text" />
<select name="hometeam_position">
<option value="<?php echo $hometeam_position ?>"><?php echo $hometeam_position ?></option>
<option disabled="disabled" value="...">...</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">3</option>
<option value="5">5</option>
<input name="submit" type="submit" value="UPDATE" />
</form>

您将使用表单操作重定向到执行更新的脚本。在这个脚本中,您可以使用$_POST数组访问表单输入元素。至于如何执行更新查询,一个示例可以是:

$query="UPDATE mytable 
        SET title = '".$title."', name = '".$name."', date = '".$date."' 
        WHERE id = ".$id;
$result=mysql_query($query);
更新:

脚本的一个例子可以是:

$hometeam_position = $_POST['hometeam_position']; //access the selected option when submitting
$countdown_module = $_POST['countdown_module']; //access the text input
$query = "UPDATE jos_gm_nextmatch SET countdown_module = '".$countdown_module."', hometeam_position = '".$hometeam_position."' WHERE id = 1";
$result=mysql_query($query);

您可以在从数据库中选择字段之前或之后简单地增加它们

...
if (isset($_POST['submit'])) {
    $stmt = "UPDATE jos_gm_nextmatch
             SET countdown_module = " . $_POST['countdown_module'] .
           " , hometeam_position =" . $_POST['hometeam_position'] . 
           " WHERE id=1";
    mysql_query($stmt);
}
mysql_close();
相关文章: