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我一直在尝试使用一个简单的 POST 请求向 MySQL 数据库添加一些基本信息，但我似乎运气不佳 - 它不会将元组添加到数据库中。我知道这段代码运行它请求的 PHP 文件，就好像我告诉 PHP 文件将虚拟信息添加到它工作正常的数据库一样，但是 POST 字符串中的信息没有被添加。

这是我的Java文件：

public class Test {

public static void main(String[] args) {

    int id = 3696;
    String name = "David";
    int steps = 6989;

    String strData = encode(id, name, steps);
    submit(strData);
}

public static String encode(int id, String name, int steps) {

    String strToSend;

    try
    {
        strToSend = URLEncoder.encode("id", "UTF-8") + "=";
        strToSend += URLEncoder.encode("" + id, "UTF-8");
        strToSend += "&" + URLEncoder.encode("name", "UTF-8") + "=";
        strToSend += URLEncoder.encode(name, "UTF-8");
        strToSend += "&" + URLEncoder.encode("steps", "UTF-8") + "=";
        strToSend += URLEncoder.encode("" + steps, "UTF-8");
    }

    catch (Exception e) {
        return "Error encoding data for submission";
    }

    return strToSend;
}

public static String submit(String strToSend) {

    URL urlDest;
    String strData = strToSend;
    System.out.println(strData);
    HttpURLConnection ucConnect;

    try 
    {
        urlDest = new URL("http://localhost/~David/posttest.php");
    }
    catch (Exception e)
    {
        return "Error in submitting entry";
    }


    try
    {
        ucConnect = (HttpURLConnection) urlDest.openConnection();
        ucConnect.setRequestProperty("Content-type","application/x-www-form-urlencoded");
        ucConnect.setRequestMethod("POST");
        ucConnect.setDoOutput(true);

    }

    catch (Exception e)
    {
        return "Error in connecting to URL";
    }

    try
    {
        OutputStreamWriter streamWrite = new OutputStreamWriter(ucConnect.getOutputStream());
        streamWrite.write(strData);
        System.out.println(streamWrite.toString());
        streamWrite.flush();

        //Get the response

        ucConnect.getInputStream();
        System.out.println(ucConnect);
        streamWrite.close();
    }

    catch (Exception e) 
    {
        return "Problem";
    }

    System.out.println("SUCCESS");
    return "Yay!";
}

}

这是 posttest.php

<?php

$con=mysqli_connect(removed);

//check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}


$id = $_POST["id"];
$strName = $_POST["name"];
$noSteps = $_POST["steps"];

mysqli_query ($con, "INSERT INTO People VALUES ($id, $strName, $noSteps)" || die();

mysqli_close($con);
echo json_encode($arr);

?>

我发送的字符串的打印格式是：

id=3696&name=David&steps=6989

谢谢您的帮助。

4

1 回答 1

1

线

mysqli_query ($con, "INSERT INTO People VALUES ($id, $strName, $noSteps)" 
  || die();

语法不正确。请在 mysqli_query 末尾包含一个 )：

mysqli_query ($con, "INSERT INTO People VALUES ($id, $strName, $noSteps)" ) 
  || die();
于 2013-10-30T23:26:11.650 回答