我有2张桌子。 我想将一些sql条目传输到另一个表。 13是表1中的catid,但在表2中我的catid是不同的。
table1 table2
catid title alias catid title alias
13 abc xyz 78 abc xyz
13 test test 78 test test
答案 0 :(得分:0)
您可以将其传递给静态,请执行以下操作: -
INSERT INTO table1 (catid, title, alias) SELECT '2', title, alias FROM table2
可选,如果您需要where子句
WHERE catid ='13'
答案 1 :(得分:-1)
Use This Sql Query:
For Getting the data from table1:
$sql = "SELECT * FROM table1 WHERE catid=13";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$title=$row["title"];
$alias=$row["alias"];
}
} else {
echo "0 results";
}
For Inserting the data into table2:
update table table2 set title='$title', alias='$alias' where catid=78
答案 2 :(得分:-1)
Use This Sql Query:
For Getting the data from table1:
$sql = "SELECT * FROM table1 WHERE catid=13";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$catid=$row["catid"];
$title=$row["title"];
$alias=$row["alias"];
}
} else {
echo "0 results";
}
For Inserting the data into table2:
update table table2 set catid='$catid', title='$title', alias='$alias' where catid=78