Question

我使用php变量动态创建了以下json，它给了我：

{"message":"invalid json"}

代码：

$json = '{
"sex": "'.$gender.'",
"age": '.$age.',
"evidence": [
{"id": "'.$symptom_id1.'", "choice_id": "present", "initial": true}, 
{"id": "'.$symptom_id2.'", "choice_id": "present", "initial": true},
{"id": "p_7", "choice_id": "'.$test1.'"},
{"id": "p_8", "choice_id": "'.$test2.'"}, 
{"id": "p_9", "choice_id": "'.$test3.'"},
{"id": "p_10", "choice_id": "'.$test4.'"},
{"id": "p_28", "choice_id": "'.$test5.'"},
{"id": "'.$location.'", "choice_id": "present"}
],
"extras": {
"disable_groups": true
}
}';
$ch = curl_init('https://api.infermedica.com/v2/diagnosis');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");                                                                     
curl_setopt($ch, CURLOPT_POSTFIELDS, $json);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE); 
curl_setopt($ch, CURLOPT_HTTPHEADER, [
'Content-Type: application/json',                                                                                
'Content-Length: ' . strlen($json),
'app_id: '. APP_ID,
'app_key: '. APP_KEY,
'Interview-Id: '.$interview_id,
'User-Id: '. $hasheduser_id
]);                                                                                                                   
$result = curl_exec($ch);

echo '<pre>';
print_r( $result );
echo '</pre>';

对值进行硬编码以代替变量可以正常工作。作为一个新手，我无法理解为什么它没有良好的形成。

请求帮助。

Answer 1

您永远不应该尝试自己创建有效的JSON，它只会导致安全问题和错误，而是使用数据结构并将其传递给public Form1() { InitializeComponent(); getAvailablePorts(); } void getAvailablePorts() { // Get a list of existing ports String[] ports = SerialPort.GetPortNames(); String[] baudrates = new String[] { "9600", "115200" }; // Insert the lists into the combo boxes comboBox1.Items.AddRange(ports); comboBox2.Items.AddRange(baudrates); // Select the default setting if at least a com port is found. if (ports.Count() != 0) { comboBox1.SelectedIndex = 0; comboBox2.SelectedIndex = 1; } } private void button1_Click(object sender, EventArgs e) { try { if (comboBox1.Text == "" || comboBox2.Text == "") { textBox1.Text = "Please select port settings"; } else { serialPort1.PortName = comboBox1.Text; serialPort1.BaudRate = Convert.ToInt32(comboBox2.Text); serialPort1.DataReceived += new SerialDataReceivedEventHandler(serialPort1_DataReceived); serialPort1.Open(); progressBar1.Value = 100; button1.Enabled = false; button2.Enabled = true; } } catch (UnauthorizedAccessException) { textBox1.Text = "UnauthorizedAccess"; } } private void serialPort1_DataReceived(object sender, SerialDataReceivedEventArgs e) { SerialPort sp = (SerialPort)sender; string value = sp.ReadExisting(); processValue(value); } //Delegate to update received data in the textbox private delegate void processValueDelegate(string value); private void processValue(string value) { if (this.InvokeRequired) { this.BeginInvoke(new processValueDelegate(processValue), new object[] { value }); } else { // ... update GUI in here ... textBox1.AppendText(value); DataExtraction(value); } } string UnreadBuffer = ""; private void DataExtraction(string RxString) { UnreadBuffer += RxString; // Search for Data Header "<=>" int start = UnreadBuffer.IndexOf("<=>"); int delimiter = UnreadBuffer.IndexOf("#"); if (start >= 0) { string row = UnreadBuffer.Substring(start); if (delimiter >= 0) { string[] ExtractedDataFields = row.Split('#'); if (ExtractedDataFields.Length >= 7) { textBox2.Text = string.Format("{0:HH:mm:ss tt}", DateTime.Now); textBox3.Text = ExtractedDataFields[1]; textBox4.Text = ExtractedDataFields[2]; textBox5.Text = ExtractedDataFields[3]; textBox6.Text = ExtractedDataFields[5]; UnreadBuffer = ""; } } } }。

您使用的示例数据结构：

json_encode

然后编码：

$to_json = [
  'sex' => $gender,
  'age' => $age,
  'evidence' => [
    ['id' => $symptom_id1, 'choice_id' => 'present', 'initial' => true],
    ['id' => $symptom_id2, 'choice_id' => 'present', 'initial' => true],
    ['id' => 'p_7', 'choice_id' => $test_1],
    ['id' => 'p_8', 'choice_id' => $test_2],
    ['id' => 'p_9', 'choice_id' => $test_3],
    ['id' => 'p_10', 'choice_id' => $test_4],
    ['id' => 'p_28', 'choice_id' => $test_5],
    ['id' => $location, 'choice_id' => 'present'],
  ],
  'extras' => [
    'disabled_groups' => true
  ],
];

希望这有帮助。

Answer 2

由于变量第二行有拼写错误，您的JSON无效。

"age": '.$age.',

应该是

"age": "'.$age.'",

或者什么会更好

"age": "$age",

但是，完成您尝试做的事情的最佳方法是创建一个PHP数组，并将其传递给json_encode。

$array = ['hello, i am a key'=>'i am some data'];
var_dump(json_encode($array));

json里面的php变量会导致json无效

2 个答案: