我正在数据库的两个表中插入数据。我使用了LAST_INSERT_ID()
函数
但我收到此错误
PHP Fatal error: Call to undefined function LAST_INSERT_ID()
这是我的代码:
$sql = "";
$sql .= "BEGIN; ";
$sql .= "INSERT INTO circle_call_prefixes (";
$sql .= " circle";
$sql .= " ,prefix";
$sql .= " ) VALUES (";
$sql .= " :cid";
$sql .= " ,:prefix";
$sql = "INSERT INTO circle_call_destinations (";
$sql .= " autoNo";
$sql .= " ,destination";
$sql .= " ,source_circle";
$sql .= " ) VALUES (";
$sql .= LAST_INSERT_ID();
$sql .= " ,:prefix";
$sql .= " ,:prefix";
$sql .= " COMMIT; ";
$sql .= " );";
$stmt = $dbh->prepare($sql);
foreach($Insert_array as $e1)
{
$stmt->bindValue(':circle', $e1['cid']);
$stmt->bindValue(':prefix', $e1['prefix']);
$stmt->bindValue(':comment', $e1['comment']);
$stmt->bindValue(':cid', $Cid);
$stmt->execute();
}
谢谢
最佳答案
LAST_INSERT_ID
位于 SQL 中,请尝试以下操作:
$sql = "";
$sql .= "BEGIN; ";
$sql .= "INSERT INTO circle_call_prefixes (";
$sql .= " circle";
$sql .= " ,prefix";
$sql .= " ) VALUES (";
$sql .= " :cid";
$sql .= " ,:prefix";
$sql = "INSERT INTO circle_call_destinations (";
$sql .= " autoNo";
$sql .= " ,destination";
$sql .= " ,source_circle";
$sql .= " ) VALUES (";
$sql .= "LAST_INSERT_ID()";
$sql .= " ,:prefix";
$sql .= " ,:prefix";
$sql .= " COMMIT; ";
$sql .= " );";
$stmt = $dbh->prepare($sql);
foreach($Insert_array as $e1){
$stmt->bindValue(':circle', $e1['cid']);
$stmt->bindValue(':prefix', $e1['prefix']);
$stmt->bindValue(':comment', $e1['comment']);
$stmt->bindValue(':cid', $Cid);
$stmt->execute();
}
关于PHP fatal error : Call to undefined function LAST_INSERT_ID(),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36028098/