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();