Question

我正在尝试构建一个非常基本的API，我有一个从MySQL视图中提取数据的查询，我可以回答查询，因为json没有问题，但我想把查询放入一个函数，以便能够从API脚本中调用它...我只是在将代码放入函数时遇到了一些麻烦。

这是有效的颂歌.....

<?php
$servername = "database.com";
$username = "username";
$password = "password";
$dbname = "db1";

// Create connection
$conn = new mysqli($servername, $username, $password,          $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo $row["year"]. " - " . $row["Month"]. " " . $row["the_days"]. "<br>";
}
} else {
echo "0 results";
}


$conn->close();
?>

...这是我试图把它变成一个函数（它不起作用！）......

//Connection as above
function available_dates() {
$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row

$encodeArray = array();

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row;
}
} else {
echo "0 results";
}
$dates = array();
$dates = json_encode($encodeArray);
return $dates;
}

available_dates();

$conn->close();
?>

我刚刚开始使用各种功能，所以我希望我的错误很滑稽！ ...我需要使用echo来调用函数吗？

Answer 1

你对函数的返回值一无所知。

您可以将其回显或将其放入变量中。

var myObj = GetMyObjInstance();

// Working, but apparently it's not good practise to call .scope() on an element.
myObj.onUpdated = function(){
    console.log("myObj updated");
    var v = myObj.getValue();
    var controllerDiv = document.getElementById("controller");
    var $scope = angular.element(controllerDiv).scope();

    $scope.apply(function(){
       $scope.someValue = v; 
    });
}

// Tried to do this, thinking i would get closure on the scope.
angular.module('myApp', []).controller('controller', function($scope){
   myObj.onUpdated = function(){
       console.log("myObj updated"); // Gets logged to console...
       var v = myObj.getValue();
       $scope.somevalue = v; // ... but somevalue does not get displayed. 
       $scope.apply(); // Error, says it's not a function, so maybe this is not the right object?.
   } 
});

或

echo available_dates();

或

print_r(available_dates());

Answer 2

你没有返回任何东西，你应该为某些东西指定返回值。你应该像这样调用你的函数。

function available_dates() {
$sql = "SELECT * from DB_Available_Dates";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row

$encodeArray = array();

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row;
}
} else {
echo "0 results";
}
$dates = array();
$dates = json_encode($encodeArray);
return $dates;
}

$result=available_dates();

Answer 3

您在代码中返回一个数组：

$dates = array();
$dates = json_encode($encodeArray);
return $dates;

您应该像这样回复json_encode：

echo json_encode($encodeArray);

然后处理格式化/显示（或者，您可以像在第一个示例中那样进行回显并回显准备显示的输出）：

while($row = $result->fetch_assoc()) {
    $encodeArray[] = $row["year"]. " - " . $row["Month"]. " " . row["the_days"].<br>";
}

然后

echo json_encode($encodeArray);

PHP将MySQL转化为json - 试图将查询转化为函数

3 个答案: