php - 如何在 postgresql 查询中使用 php 变量

Question

我有两个 php 变量，一个是整数，另一个是 json，我将它们转换成字符串变量，然后将它们插入到 postgresql 数据库中。

将整数转换为字符串变量:

$string1 = (string)$integer;

将来自 facebook api 的 json 转换为字符串变量:

$string2 = json_encode($json);

现在，我必须将这两个字符串变量插入到 postgres 数据库中:

$query = "INSERT INTO interests VALUES(". $string1 ." ," . $string2 .")"; 
pg_query($con, $query) or die("Cannot execute query: $query\n");

这是行不通的。我尝试了很多解决方案，但仍然无法正常工作。

我更改了我的函数以插入到数据库中

function push_interests(){
$id = $facebook->getUser();
$int = $facebook->api('/me/interests');     
$host = "hostname"; 
$user = "user"; 
$pass = "password"; 
$db = "database"; 

$con = pg_connect("host=$host dbname=$db user=$user password=$pass")
    or die ("Could not connect to server\n"); 


$id = (string)$id;
$int = json_encode($int);

$sql = "INSERT INTO interests VALUES($1,$2)"; 
pg_prepare($con,'my_insert', $sql) or die ("Cannot prepare statement1\n") ;
pg_execute($con,'my_insert', array($id,$int)) or die ("Cannot execute statement1\n");

pg_close($con); 

}

输出是:无法执行statement1我创建了如下数据库:

$query = "DROP TABLE IF EXISTS interests"; 
pg_query($con, $query) or die("Cannot execute query: $query\n");

$query = "CREATE TABLE interests(id VARCHAR(25) PRIMARY KEY, interests VARCHAR(500))";  
pg_query($con, $query) or die("Cannot execute query: $query\n"); 

Answer 1

因为字符串需要用简单的引号括起来。我强烈建议您使用准备好的语句来忽略此类问题并确保正确的变量转义以防止您的应用程序通过 SQL 注入(inject)被黑客攻击。

$sql = "INSERT INTO interests VALUES ($1, $2)";
$result = pg_prepare($con, 'my_insert', $sql);
$result = pg_execute($con, 'my_insert', array($string1, $string2));

参见 http://php.net/manual/en/function.pg-prepare.php

编辑:这是我测试过的实际代码:

<?php

$con = pg_connect('')
    or die ("Could not connect to server\n"); 

$id = (string) 5;
$int = json_encode(array('pika' => 'chu', 'plop' => array(1, 2, 3)));

$query = "CREATE TABLE interests(id VARCHAR(25) PRIMARY KEY, interests VARCHAR(500))";
pg_query($query) or die('creating table failed.');

$sql = "INSERT INTO interests (id, interests) VALUES ($1, $2)";
pg_prepare('my_query', $sql);
pg_execute('my_query', array($id, $int)) or die("Error while inserting.");