PHP Function List
mysql_insert_id. doesn't exist. Closest matches:
- mysql_insert_id
- fbsql_insert_id
- mysql_list_dbs
- mysql_list_fields
- mysql_select_db
- maxdb_insert_id
- msql_list_dbs
- msql_list_fields
- mysqli_set_opt
- mysql_fetch_field
- msql_select_db
- sqlite_last_insert_rowid
- db2_last_insert_id
- mysql_thread_id
- mysql_create_db
- mssql_select_db
- cubrid_insert_id
- mysql_num_fields
- msql_fetch_field
- mysql_connect
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mysql_insert_id() 返回给定的 link_identifier 中上一步 INSERT 查询中产生的 AUTO_INCREMENT 的 ID 号。如果没有指定 link_identifier,则使用上一个打开的连接。 如果上一查询没有产生 AUTO_INCREMENT 的 ...
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mysql_insert_id() は直近のクエリに対して働くので、 値を生成したクエリの直後に mysql_insert_id() をコールすることを忘れないようにしてください。 注意:
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if( strtolower(substr($query,0,6)) != 'select' ) return array(mysql_affected_rows(),mysql_insert_id()); $count = @mysql_num_rows($r); if( !$count ) return 0; if( $count == 1 ) { if( $assoc ) $f = mysql_fetch_assoc($r);
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mysql_insert_id — 取得上一步 INSERT 操作产生的 ID mysql_list_dbs — 列出 MySQL 服务器中所有的数据库 mysql_list_fields — 列出 MySQL 结果中的字段 mysql_list_processes — 列出 MySQL 进程
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Note that if you Call a MySQL stored procedure to insert a new record and then reference $db->insert_id; you will get 0 back, not the last inserted ID. It is therefore necessary to add a line to your MySQL Stored Procedure such as
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See Also mysql_affected_rows() - Get number of affected rows in previous MySQL operation mysql_insert_id() - Get the ID generated in the last query mysql_stat() - Get current system status
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But you can retrieve the last inserted Id executing a query asking for the function LAST_INSERT_ID() (at least in MySQL) Try this: ($o_db is the declared adapter) $last_id = $o_db->fetchAll('SELECT LAST_INSERT_ID() as last_id');
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// finally insert the new record - voila - 4 lines! $id_max = mysql_result(mysql_query("SELECT MAX(id) FROM table_name"),0,0) or die("Could not execute query"); $entity = mysql_fetch_array(mysql_query("SELECT * FROM table."
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$insert = mysql_query("SELECT `test1`.*, `test2`.*, FROM `test1`, `test2`, ` WHERE ((`test1`.`author` = `test2`.`ID`) AND (`test2`.`ID` ='$number')) ORDER BY `toetsen`.`autoID` DESC",$link); echo mysql_result( $insert , 0, 'test1.question') ;
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mysql_insert_id mysql_get_server_info ... This causes problems when you want to use the result of the update to determine if there's need to do an INSERT. With MySQL you ...

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