我想,我从MySQL得到我的PHP阵列 谁能帮助我的结果转换的MySQL结果到PHP数组
解决<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "women";
$conn = new mysqli($servername, $username, $password, $dbname);
$id=$_GET['id'];
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT DAY(ADDDATE(`dateDebutC`, `dureeC`)) AS MONTHS,
DAY(ADDDATE(ADDDATE(`dateDebutC`, `dureeC`),`dureeR`))AS DAYS
FROM normalW
where id = '$id'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
foreach($new_array as $array){
echo $row['DAYS'].'<br />';
echo $row['MONTHS'].'<br />';
}
} else {
echo "0 results";
}
$conn->close();
?>
问题谢谢你们
你从来没有设置'$ new_array' ...获取'$ result'。您还应该参数化该查询,您可以按照原样打开注射。 – chris85
你的脚本存在[SQL注入攻击]的风险(http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) 看看发生了什么[小鲍比表](http://bobby-tables.com/)即使 [如果你是逃避投入,它不安全!](http://stackoverflow.com/questions/5741187/sql-injection-that-gets -around-mysql-real-escape-string) 使用[prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) – RiggsFolly
你可以举一个你需要的例子请结果。现在它不是很清楚你想要做什么 – RiggsFolly