我正在开发一个web服务和android应用程序web服务是在PHP从MySQL数据库获取数据...我遇到的问题是,一些项目有多个细节,但我的代码是只为每个项目提供一个细节。我以JSON格式获取数据。嵌套的while循环不工作
您可以检查此Here
这里是我的代码提前
function requiredData(){
$db = $this->dbConnection();
//$sql = "SELECT * FROM projects JOIN project_details ON projects.project_id=project_details.project_id";
$sql = "SELECT * FROM projects";
$queryResult = $db->query($sql);
if($queryResult->num_rows > 0){
while($row = $queryResult->fetch_assoc()){
$pid = $row['project_id'];
$detailsql = "SELECT * FROM project_details WHERE project_id=$pid";
$sqlResult = $db->query($detailsql);
if($sqlResult->num_rows > 0){
while ($d = $sqlResult->fetch_assoc()){
$r = array(
"project_id" => $d['project_id'],
"project_detail" => array(
"work_done" => $d['project_detail'],
"payment_for_work" => $d['project_payment'],
"payment_status" => $d['project_payment_status'],
"detail_id" => $d['project_detail_id']
)
);
}
}
$results[$row['project_name']] = array(
"project_id" => $row["project_id"],
"project_start_date" => $row["project_start_date"],
"project_due_date" => $row["project_due_date"],
"project_currency" => $row["project_currency"],
"project_work_details" => $r
);
}
}
return $results;
}
感谢您的帮助
什么?我检查了你指定的链接,它显示了多个项目。你到底想要什么? – Umair
一些项目有多个project_details ...但目前有一个显示...例如项目ID 20有多个project_detail数组,,,但只有一个显示出来... – FaISalBLiNK