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所以我在一个php页面中有一个表格：

$page .='<form method="POST" action="delete.php" ID="orgForm">';
$page .= "<table> \n";
$page .= "<tr>\n";
//Decides what to display based on logged in. Only if logged in can you see all the contact's info
$page .= "<th>ID</th> \n <th>First Name</th> \n <th>Last Name</th> \n <th>Phone Number</th> \n <th>Email</th> \n";
//Loops through each contact, displaying information in a table according to login status
$sql2="SELECT cID, firstName, lastName, phoneNum, email FROM Contact WHERE oID=".$_GET['orgID'];
$result2=mysql_query($sql2, $connection) or die($sql1);
while($row2 = mysql_fetch_object($result2))
{
    $page .= "<tr>\n";
    $page .= "<td>".$row2->cID."</td>\n";
    $page .= "<td>".$row2->firstName."</td>\n";
    $page .= "<td>".$row2->lastName."</td>\n";
    $page .= "<td>".$row2->phoneNum."</td>\n";
    $page .= "<td>".$row2->email."</td>\n";
    $page .= '<td><input type="checkbox" name="checkedItem[]" value="'.$row2->cID.'"></input></td>'."\n";
    $page .="</tr>";
}
$page .= '<input name="deleteContacts" type="submit" value="Delete Selected Contacts" />'."\n";
$page .= "</form>\n";

$page .='<script src="assets/js/orgDetails.js" type="text/javascript"></script>'."\n";

我需要以某种方式在 orgDetails.js 中编写一个 jquery 脚本，当我按下删除按钮时，它能够删除选中的行。更改必须在不刷新的情况下出现在屏幕上，而且我还需要能够从 sql db 中删除实际行。有人可以帮帮我吗？谢谢。

4

1 回答 1

1

In action url delete.php, After submit this post:

if ($_POST != array()) {
    foreach ($_POST['checkedItem'] as $id) {
        mysql_query('delete from Contact where cID='.$id);
    }

    echo 'Records deleted.';
}

If you dont want page refresh when delete records:

Add to html:

<button class="delete_button">Delete selected records</button>

Add your js file:

$('.delete_button').click(function () {
    $form = $('#orgForm');

    delete_ids = [];

    $form.find('input[name=checkedItem]').each(function () {
        $checkbox = $(this);

        if ($checkbox.is(':checked')) {
            delete_ids.push($checkbox.val());
        }
    );

    $.ajax({
        url: 'delete.php',
        type: 'post',
        data: {delete_ids: delete_ids},
        success: function (result_html) { alert(result_html); },
    });
});

And in delete.php:

if ($_POST != array()) {
    foreach ($_POST['delete_ids'] as $id) {
        mysql_query('delete from Contact where cID='.$id);
    }

    echo 'Records deleted.';
}
于 2013-02-13T02:04:03.630 回答