android java文件。
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("name", username));
params.add(new BasicNameValuePair("channel", channel));
// getting details by making HTTP request
JSONObject json = jParser.makeHttpRequest(
"localhost/users/url_delete_users.php", "POST", params);
url_delete_users.php
if (isset($_POST['name']) && isset($_POST['channel'])) {
$name = $_POST['name']; $channel = $_POST['channel'];
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// mysql update row with matched pid
$result = mysql_query("DELETE FROM users WHERE name = $name AND channel=$channel");**
mysql_query有什么问题吗? 它不起作用...... (create db php source运行良好。只有删除获取错误)
答案 0 :(得分:1)
确保上面的一切正常,然后试试这个:
$result = mysql_query("DELETE FROM users WHERE name = '".$name."' AND channel='".$channel."'");
答案 1 :(得分:0)
你能试试吗,
$result = mysql_query("DELETE FROM users WHERE
`name` = '".$name."' AND channel='".$channel."' ")
or die(mysql_error());
而不是
$result = mysql_query("DELETE FROM users WHERE name = $name AND channel=$channel");
答案 2 :(得分:0)
试试这个
$result = mysql_query("DELETE FROM users WHERE name = '$name' AND channel='$channel'");
答案 3 :(得分:0)
我认为
$result = mysql_query("DELETE FROM users WHERE name = $name AND channel=$channel");
应该是
$result = mysql_query("DELETE FROM users WHERE name = '$name' AND channel='$channel'");