[英]How to select the last inserted ID on concatenated values
我正在嘗試獲取多個插入行的最后一個插入ID。
record_id是自動遞增
$sql = "INSERT INTO records (record_id, user_id, status, x) values ";
$varray = array();
$rid = $row['record_id'];
$uid = $row['user_name'];
$status = $row['status'];
$x = $row['x'];
$varray[] = "('$rid', '$uid', '$status', '$x')";
$sql .= implode(',', $varray);
mysql_query($sql);
$sql2 = "INSERT INTO status_logs (id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES";
$varray2[] = "(' ', mysql_insert_id(), '$status', '$uid', '$x')";
$sql2 .= implode(',', $varray2);
mysql_query($sql2);
結果如下:
INSERT INTO records (record_id, user_id, notes, x) values ('', '1237615', 'this is a note', 'active')
INSERT INTO status_logs (log_id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES('', INSERT INTO records (record_id, user_id, notes, x) values ('', '1237615', 'this is a note', 'active')
INSERT INTO status_logs (log_id, record_id, status_id, date, timestamp, notes, user_id, x) VALUES('', mysql_insert_id(), '1', '2013:05:16 00:00:01', '', this is a note'', '1237615', 'active'), '1', '2013:05:16 00:00:01', '', this is a note'', '1237615', 'active')
mysql_insert_id()
沒有值。
除非我誤讀了您的代碼,否則您是在SQL內調用PHP函數mysql_insert_id
?
您需要做的是先將其捕獲到PHP變量中,然后在SQL中使用該變量。 像這樣:
// Run the first query
mysql_query($sql);
// Grab the newly created record_id
$recordid= mysql_insert_id();
然后在第二個INSERT中使用:
$varray2[] = "(' ', $recordid, '$status', '$uid', '$x')";
您正在混合php函數mysql_insert_id()
和SQL INSERT
語句語法。
無論是使用MySQL函數LAST_INSERT_ID()
在VALUES
子句INSERT
語句
INSERT INTO records (user_id, notes, x) VALUES('1237615', 'this is a note', 'active');
INSERT INTO status_logs (record_id, status_id, date, timestamp, notes, user_id, x)
VALUES(LAST_INSERT_ID(), '1', ...);
^^^^^^^^^^^^^^^^^
或通過在第一個mysql_query()
之后立即單獨調用mysql_insert_id()
來檢索最后插入的ID。 然后在您將其用作第二個查詢的參數時使用該值。
$sql = "INSERT INTO records (user_id, ...)
VALUES(...)";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error()); //TODO beter error handling
}
$last_id = mysql_insert_id();
// ^^^^^^^^^^^^^^^^^^
$sql2 = "INSERT INTO status_logs (record_id, ...)
VALUES $last_id, ...)";
$result = mysql_query($sql);
if (!$result) {
die('Invalid query: ' . mysql_error()); //TODO beter error handling
}
注意:
附帶說明 :與其插入查詢字符串,而不是讓它對sql-injections開放, mysqli_*
考慮將准備好的語句與mysqli_*
或PDO
。
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