如何搜索数据库列和 return 匹配行

Question

我有以下数据库（名为“Account_info”）：

    |id (All Unique)|domain_name (All unique)|   account_name |account_key|
    |---------------|------------------------|----------------|-----------|
    | 1             |example.com             |account_23657612|889977     |
    | 2             |example.net             |account_53357945|889977     |
    | 3             |example.edu             |account_53357945|889977     |
    | 4             |example.xyz             |account_93713441|998822     |

我希望在该数据库中搜索域“example.net”，并将该域的“帐户密钥”列作为变量返回。

示例：

SEARCH TERM: example.net

RESPONSE: 889977

到目前为止我的代码：

$domainSearch = "example.net";
$sql = mysqli_query($connect,"SELECT `account_name` FROM `Account_info` WHERE `domain_name`='$domainSearch'");
if($sql){
    //Some code here that sets the variable "result" to "889977"
}

Answer 1

要获取帐户密钥 return，只需从结果集中获取数据并将其return作为关联数组

另一方面，请使用参数化parpared语句来避免SQL注入攻击。

因此改变

$sql = mysqli_query($connect,"SELECT `account_name` FROM `Account_info` WHERE `domain_name`='$domainSearch'");

至


$sql = "SELECT * FROM Account_info WHERE domain_name=?"; // SQL with parameters
$stmt = $connect->prepare($sql); 
$stmt->bind_param("s", $domainSearch);
$stmt->execute();
$result = $stmt->get_result(); // get the mysqli result
$user = $result->fetch_assoc(); // fetch data   

echo $user["account_key"];

// to assign as a variable -- $var1=$user["account_key"];

如何搜索数据库列和 return 匹配行

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