使用php

Question

我试图为Postgres数据库编写SELECT查询但不是我收到了一些错误 - pg_query（）：查询失败：错误：语法错误在或附近＆＃34; $＆＃34;第1行

$findResult = pg_query('SELECT * FROM fingerprintdata WHERE buildingname = $building  ');
$numrows    = pg_num_rows($findResult);

Answer 1

查询应该是这样的：

pg_query('SELECT * FROM fingerprintdata WHERE buildingname = "$building"  ');

变量应该在引号内＆＃34;＆＃34;

Answer 2

$findResult = pg_query($conn, 'SELECT * FROM fingerprintdata WHERE buildingname = "$building"');
$numrows = pg_num_rows($findResult);

Answer 3

请尝试以下解决方案。你没有在单引号中加上等号，这就是你得到变量错误的原因。

$findResult = pg_query('SELECT * FROM fingerprintdata WHERE buildingname =' . $building);
$numrows    = pg_num_rows($findResult);

Answer 4

$query = 'SELECT * FROM fingerprintdata WHERE buildingname = '.pg_escape_literal($building); // to be able to var_dump($query) to see, what it actually looks like
$findResult = pg_query($query);

在您的情况下，您使用的是单引号，但不会在内部扩展$building。如果您使用双引号，它会起作用，但您很容易受到SQL注入（例如$building = "xxx'; DROP DATABASE ...";）