我看了看,但还是不明白。
我的问题是:
数据库有5个表,分别命名为pbtest01、pbtest02、pbtest03、pbtest04和pbtest05。我使用以下代码循环表:
$x = 3;
for($k = 1; $k <= $x; $k++){
$sql5 = "SELECT *
FROM CONCAT('pbtest0',$k)
WHERE id = '930820105627'
";
$data5 = mysql_query($sql5) or die(mysql_error().$sql5);
$list5 = mysql_fetch_array($data5);
$var[$k] = $sql5['value'];
}
echo $var[1];
echo $var[2];
echo $var[3];
但得到以下错误:
You have an error in your SQL syntax; check the manual that corresponds to
your MySQL server version for the right syntax to use near
'CONCAT('pbtest0',1) WHERE id = '930820105627'' at line 1
SELECT *
FROM CONCAT('pbtest0',1)
WHERE id = '930820105627'
有人能帮我吗?
最佳答案
试试这个:
for($k = 1; $k <= $x; $k++){
$sql5 = "SELECT *
FROM pbtest0" . $k . "
WHERE id = '930820105627'
";
...
}
CONCAT()
是一个MySQL函数,需要用PHP构建表名。