如何获取表中存储的SQL查询并为表中的所有查询创建动态PHP页面

Question

I'm trying to retrieve dynamic query PHP pages. First page passing values of city or id next page by post method.

Table looks like this:

TABLE1

-----------------------------------------------------
1     select * from tablename1 where ID = $id
2     select * from tablename1 where city = $city
-----------------------------------------------------

Now I run page contain code:

    $id = $_POST["id"];
    $city= $_POST["city"];
    $tabledata = mysql_query("SELECT * FROM table1 ");
    $a=mysql_fetch_row($tabledata );

    $query_table=$a['1']; // retrieve sql query from table1 which stored in 2nd column
    echo " $query_table"; // display query
    $result = mysql_query($query_table);

    while ($field = mysql_fetch_row( $result) )
    {
        echo "<td> $field[1] </td>";
    }

Now, the problem is that I can't get results. $result is empty?

How pass $city or $id value to query?

But if I remove WHERE condition from TABKE1 then it works well. So, how can I make for WHERE condition to pass parameters?

how can i do this . any other option here to call stored query on 1 table and process it on other page and oher table

Answer 1

在第一个语句中使用WHERE和OR子句，然后可以迭代作为行返回的每个结果，并且可以避免上面的复杂方法。

$tabledata = mysql_query("SELECT * FROM table1 WHERE ID = {$id} OR city = {$city}");