Question

<?php 
$successreturn[]=array(
                    "id"=>"any",
                    "firstname"=>"any",
                    "lastname"=>"any",
                    "dateofbirth"=>"any",
                    "city"=>"any",
                    "gender"=>"any"
                );
header("Access-Control-Allow-Origin: *");
$servername="localhost";
$username="root";
$password="sandeepchetikam";
$dbase="mydb";

$conn=mysqli_connect($servername,$username,$password,$dbase);
if (!$conn) {
    echo "Connection Problem".mysqli_connect_error($conn);
}

$sql= "SELECT * FROM Employees";
$result = mysqli_query($conn,$sql);
$count = mysqli_num_rows($result);
$value=0;

if(!$result){
    echo "Connection Failed " .mysqli_connect_error($result);
}

while($row = mysqli_fetch_assoc($result)){

    $successreturn[$value]['id']            =$row['id'];
    $successreturn[$value]['firstname']     =$row['firstname'];
    $successreturn[$value]['lastname']      =$row['lastname'];
    $successreturn[$value]['dateofbirth']   =$row['dateofbirth'];
    $successreturn[$value]['city']          =$row['city'];
    $successreturn[$value]['gender']        =$row['gender'];
    $value++;
};
echo json_encode($successreturn);
?>

输出：

[{"id":"any","firstname":"any","lastname":"any","dateofbirth":"any","city":"any","gender":"any"}]

我正在尝试将JSON值返回到我的角度服务中。但是当表中没有更多行时。它将Colomn值返回为“any”。为什么会这样？如何返回没有值的空行？

Answer 1

当没有选择行时，while循环没有执行，并且初始$ successreturn数组返回值为＆＃39;任何＆＃39; 如果没有选择行则返回没有值

只需更改此

$successreturn[]=array(
        "id"=>"any",
       "firstname"=>"any",
      "lastname"=>"any",
    "dateofbirth"=>"any",
    "city"=>"any",
    "gender"=>"any");

到

$successreturn[]=array(
        "id"=>"",
       "firstname"=>"",
      "lastname"=>"",
    "dateofbirth"=>"",
    "city"=>"",
    "gender"=>"");

修改： -

$successreturn=[];

在视图中检查值是否为空，然后显示数据，否则显示数据。我会帮助你。

Answer 2

您需要以不同的方式控制程序流程

<?php 

header("Access-Control-Allow-Origin: *");
$servername="localhost";
$username="root";
$password="sandeepchetikam";
$dbase="mydb";

$conn=mysqli_connect($servername,$username,$password,$dbase);
if (!$conn) {
    echo "Connection Problem".mysqli_connect_error($conn);
}

// first only select what you want to use from the row
$sql= "SELECT `id`,`firstname`,`lastname`,
              `dateofbirth`,`city`,`gender`
         FROM Employees";

$result = mysqli_query($conn,$sql);
if(!$result){
    // you only use `mysqli_connect_error` to get connection errors
    // use mysqli_error($result) for normal query errors
    echo "Query failed " . mysqli_error($result);
    echo json_encode(array('error'=>'Query failed'));
    exit;
}

if ( mysqli_num_rows($result) > 0 ) {
    while($row = mysqli_fetch_assoc($result)){
        // now as you only have what you want in your assoc array
        $rows[] = $row;
    }
    echo json_encode($rows);
} else {
    // no data returned from query
    // return something so the calling code knows what to do
    echo json_encode(array('error'=>'No data in table'));
}
?>

当SQL表中没有行时输出“any”的值

2 个答案: