Question

表单由一个函数构成，该函数打印出所有等待批准的行;

function all_reviews(){
$query1= mysql_query("SELECT * FROM {$table} WHERE first != '' && disp_state= '0' && rejected != '1'");
if (!$query1) {
    die("Query to show fields from table failed");
}
$fields_num = mysql_num_fields($query1);

// printing table rows
while($row = mysql_fetch_array($query1))
{
//review action buttons
        echo "<div class=\"rev_action\">";
        echo "<input type=\"hidden\" name=\"".$row['id']."\">";
        echo "<input type=\"submit\" name=\"accept\" value=\"Approve\" class=\"success rev_button\" onclick=\"document.getElementById('review_action').submit()\">";
        echo "<input type=\"submit\" name=\"reject\" value=\"Reject\" class=\"error rev_button\" onclick=\"document.getElementById('review_action').submit()\">";
        echo "</div>";
}

我猜测正确的方法是在隐藏的输入中打印出行ID。问题是，一次打印出的行可能超过1行。

如何选择与批准或拒绝按钮相关的确切行？

HTML：

<form name="form" action="include/review_submit.php" id="review_action" method="post">

    <?php all_reviews(); ?>

</form>

php（review_submit.php）：

//database server connection 
mysql_connect ("$servername","$dbusername","$dbpassword") or die(mysql_error());
//database connection 
mysql_select_db("esd_db") or die(mysql_error()); 

//data feilds from review form
if($_POST['button'] == "Approve") {
   //code to update given row with an approval status
}

//data feilds from review form
if($_POST['button'] == "Reject") {
   //code to update given row with an rejection status
}
//insert values into table
 mysql_query("UPDATE `reviews` ") or die(mysql_error());

 Print "Success"; 

?>

Answer 1

您总是可以将每一行设为不同的形式，然后将其传递给动态网址，GetDATA会告诉您选择的是哪一种。

或者，您可以不使用表单，并将每个表单链接到页面。（假设动作页面是action.php，你有100行。你可以有action.php？row = 1＆amp; ac = y，action.php？row = 1＆amp; ac = n等）

Answer 2

如果要使用基本的非AJAX表单，则应为每行添加一个复选框，并将按钮从每行移动到表单底部：

<div class="rev_action">
  <input type="checkbox" name="select_row[]" value="1">
</div>
<div class="rev_action">
  <input type="checkbox" name="select_row[]" value="2">
</div>
<!-- ... -->
<input type="submit" name="accept" value="Approve" class="success rev_button">
<input type="submit" name="reject" value="Reject" class="error rev_button">

如果您想使用ajax，则无需进行上述任何更改，并且需要为每个按钮添加点击事件：

echo "<div class=\"rev_action\">";
echo "<input class="id_input" type=\"hidden\" name=\"".$row['id']."\">";
echo "<input type=\"submit\" name=\"accept\" value=\"Approve\" class=\"success rev_button\" onclick=\"document.getElementById('review_action').submit()\">";
echo "<input type=\"submit\" name=\"reject\" value=\"Reject\" class=\"error rev_button\" onclick=\"document.getElementById('review_action').submit()\">";
echo "</div>";
...

</form>
<script>
$('input.success.rev_button').click(function(){
  var review_id = $(this).siblings('input.id_input').attr('name')
  $.ajax({
    url: '/reviews/success',
    type:'POST',
    data: {id: review_id}
  })
})
</script>

MySQL：在具有多个ID的表单中传递单行ID

2 个答案: