php - 从多个 PHP 数组创建 JSON

Question

我在 PHP 文件中创建了三个数组，所有数组都从 MYSQL 数据库获取其元素。尽管我使用了 json_encode 方法来准备数组，但结果不是可解析的 JSON。

我对 JSON 和 PHP 非常陌生，但我尝试将三个数组编码为一个变量以及三个不同的变量。但在这两种情况下，当我回显结果时，它都不是有效的 JSON。

我在 PHP 文件的开头创建了数组。

$physicianArr = array();

$patientArr = array();

$apmtArr = array();

我使用这种通用方法来填充所有数组(行用于分解三个示例，不包含在代码中。)

if (mysqli_num_rows($resultPhysician) > 0) {
  while($row=mysqli_fetch_assoc($resultPhysician)) {
    $physicianArr[] = $row;
  }
  $physicianJSON = json_encode($physicianArr);
}

if (mysqli_num_rows($resultPatient) > 0) {
  while($row=mysqli_fetch_assoc($resultPatient)) {
    $patientArr[] = $row;
  }
  $patientJSON = json_encode($patientArr);
}

if (mysqli_num_rows($resultApmt) > 0) {
  while($row=mysqli_fetch_assoc($resultApmt)) {
    $apmtArr[] = $row;
  }
  $apmtJSON = json_encode($apmtArr);
}

结果:

当我在 Web 浏览器上输入 URL 来检索此 PHP 文件时，结果是三个单独的数组而不是一个 JSON:

[
    {"UserID":"3","FirstName":"Jane","LastName":"Parkey"}, 
    {"UserID":"4","FirstName":"Jamie","LastName":"Crane"},
    {"UserID":"5","FirstName":"Jerry","LastName":"Martin"},
    {"UserID":"6","FirstName":"Alexander","LastName":"Dollar"},
    {"UserID":"8","FirstName":"Bob","LastName":"Loblaw"},
    {"UserID":"11","FirstName":"Mary","LastName":"Robbins"},
    {"UserID":"15","FirstName":"testy","LastName":"testy"}
][
    {"UserID":"1","FirstName":"Joe","LastName":"Smith"},
    {"UserID":"2","FirstName":"Adam","LastName":"Stone"},
    {"UserID":"9","FirstName":"Michael","LastName":"Jordan"},
    {"UserID":"10","FirstName":"Tom","LastName":"Holland"},
    {"UserID":"12","FirstName":"test1","LastName":"test1"},
    {"UserID":"19","FirstName":"Will","LastName":"Smith"},
    {"UserID":"20","FirstName":"Joe","LastName":"Imburgia"}
][
    {"apmtID":"1","PhysicianID":"15","apmtDate":"Jun 26, 2019","apmtTime":"09:00:00 AM","PatientID":"1"},
    {"apmtID":"2","PhysicianID":"15","apmtDate":"test","apmtTime":"test","PatientID":"1"},
    {"apmtID":"4","PhysicianID":"15","apmtDate":"Apr 20, 2019","apmtTime":"7:10:36 AM","PatientID":"1"}
]

Answer 1

这是一个建议。如果您想要一种数据结构，则只需回显一种。

// make sure all arrays exist
$physicianArr   = [];
$patientArr     = [];
$apmtArr        = [];

if (mysqli_num_rows($resultPhysician) > 0) {
    while($row=mysqli_fetch_assoc($resultPhysician)) {
        $physicianArr[] = $row;
    }
}   

if (mysqli_num_rows($resultPatient) > 0) {
    while($row=mysqli_fetch_assoc($resultPatient)) {
        $patientArr[] = $row;
    }
}

if (mysqli_num_rows($resultApmt) > 0) {
    while($row=mysqli_fetch_assoc($resultApmt)) {
        $apmtArr[] = $row;
    }
}

echo json_encode( 
                [
                    'Physicians'    => $physicianArr
                    'Patients'      => $patientArr
                    'Ampt'          => $apmtArr
                ] 
            );