Question

我一直在尝试使用简单的POST请求将一些基本信息添加到MySQL数据库中，但我似乎没有多少运气 - 它不会添加元组到数据库。我知道这段代码运行它请求的PHP文件，好像我告诉PHP文件将虚拟信息添加到数据库它工作正常，但POST字符串中的信息不会被添加。

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!";
}

}

这里是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为：

id=3696&name=David&steps=6989

感谢您的帮助。

Answer 1

该行

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

在语法上是不正确的。请在mysqli_query结尾处包括a）：

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

Java和PHP POST请求不起作用

1 个答案: