我尝试下一个脚本:
// Insert data into mysql
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES (UUID(), '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry);
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$ID')"; <--- Here is a problem
$result=mysql_query($qry2)
我不知道两个表格如何在两个表格中插入相同的UUID 。请帮我! 我非常感谢你的支持!
DONE !!! 工作脚本:
$q = "SELECT UUID() AS uid";
$res = mysql_query($q) or die('q error: '.mysql_error());
$row = mysql_fetch_assoc($res);
// Insert data into mysql
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('".$row['uid']."', '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry) or die('err 034r '.mysql_error());
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('".$row['uid']."')";
$result=mysql_query($qry2) or die('gg2345 '.mysql_error());
答案 0 :(得分:2)
在发送SELECT UUID()
之前,只需执行INSERT
并将值放入PHP中的语句中。像这样(未经测试):
$result = mysql_query("SELECT UUID() AS UUID") or die('SQL error: ' . mysql_error());
$row = mysql_fetch_assoc($result);
$UUID = $row["UUID"];
$qry="INSERT INTO $tbl_name1 (ID, REFERENCE, CODE, NAME) VALUES ('$UUID', '$REFERENCE', '$CODE', '$NAME')";
$result=mysql_query($qry);
$qry2="INSERT INTO $tbl_name2 (PRODUCT) VALUES ('$UUID ')"; <--- Here is a problem
$result=mysql_query($qry2)
另一种方法是使用用户定义的变量(see SQL Fiddle):
SET @UUID = (SELECT UUID() AS UUID);
INSERT INTO test1 VALUES(@UUID, "foo");
INSERT INTO test1 VALUES(@UUID, "bar");
答案 1 :(得分:1)
假设ID
是您可以在$qry2
之前添加的表唯一索引:
$ID = mysql_insert_id();