android/php 记录未插入 mysql

Posted 2023-04-19

【中文标题】android/php 记录未插入 mysql

【英文标题】：android/php record not inserting into mysql

【发布时间】：2013-12-31 23:17:17

【问题描述】：

我在将记录插入 mysql 表时遇到问题。只插入最后一条记录。

这是我正在处理的代码：

安卓

   String[] arrDocumentNumber, arrTransactionDate, arrItemCode;

    arrDocumentNumber = dbHelper.getAllDocumentNumbersFromOUTTRANS();
    arrTransactionDate = dbHelper.getAllTransactionDatesFromOUTTRANS();
    arrItemCode = dbHelper.getAllItemCodesFromOUTTRANS();

     protected String doInBackground(String... args) 

        // Building Parameters
        List<NameValuePair> params = new ArrayList<NameValuePair>();

        for(int i = 0; i < arrDocumentNumber.length; i++)

            params.add(new BasicNameValuePair(TAG_DOCUMENTNUMBER, arrDocumentNumber[i] ));
            params.add(new BasicNameValuePair(TAG_TRANSACTIONDATE, arrTransactionDate[i] ));
            params.add(new BasicNameValuePair(TAG_ITEMCODE, arrItemCode[i] )); 

        

        // getting JSON Object
        // Note that insert outtrans url accepts POST method
        JSONObject json = jsonParser.makeHttpRequest(url_insertTo_outtrans,
                "POST", params);

        //obj.put(params);

        //Log.d("JSON", obj.toString());

        // check log cat from response
        Log.d("Create Response", json.toString());

   

php

    <?php

/*
 * Following code will create a new product row
 * All product details are read from HTTP Post Request
 */

// array for JSON response
$response = array();

// check for required fields
if ( isset($_POST['documentnumber']) && isset($_POST['transactiondate']) && isset($_POST['itemcode']) )

    $documentnumber = $_POST['documentnumber'];
    $transactiondate = $_POST['transactiondate'];
    $locationdate = $_POST['locationdate'];
    $itemcode = $_POST['itemcode'];
    $qty = $_POST['qty'];
    $amount = $_POST['amount'];
    $reson = $_POST['reason'];
    $posted = $_POST['posted'];
    $dateposted = $_POST['dateposted'];
    $unitprice = $_POST['unitprice'];

    // include db connect class
    include('dbconnect.php');

    $result = $conn->prepare("INSERT INTO outtrans(documentnumber, transactiondate, itemcode) VALUES (:documentnumber, :transactiondate, :itemcode)");


        $result->bindParam(':documentnumber', $documentnumber);
        $result->bindParam(':transactiondate', $transactiondate);
        $result->bindParam(':itemcode', $itemcode); 
        $result->execute();



    // check if row inserted or not
    if ($result) 
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = "New record successfully created.";

        // echoing JSON response
        echo json_encode($response);
     else 
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        // echoing JSON response
        echo json_encode($response);
    
 else 
    // required field is missing
    $response["success"] = 0;
    $response["message"] = "Required field(s) is missing";

    // echoing JSON response
    echo json_encode($response);

?>

只有最后的记录被插入到 mysql 表中。我对 PHP 了解不多，所以我的 PHP 代码中可能会遗漏一些东西。我将不胜感激您的帮助。谢谢。

【问题讨论】：

能否请您发布 JSON 解析器的代码

【参考方案1】：

这是因为您只使用最后一个参数制作HttpRequest。 （即）在你的 for 循环完成后，params 将只包含最后一条记录。 所以移动

JSONObject json=jsonParser.makeHttpRequest(url_insertTo_outtrans,"POST",params);

进入 for 循环。

for(int i = 0; i < arrDocumentNumber.length; i++)

 List<NameValuePair> params = new ArrayList<NameValuePair>();
 params.add(new BasicNameValuePair(TAG_DOCUMENTNUMBER, arrDocumentNumber[i] ));
 params.add(new BasicNameValuePair(TAG_TRANSACTIONDATE, arrTransactionDate[i] ));
 params.add(new BasicNameValuePair(TAG_ITEMCODE, arrItemCode[i] )); 
 JSONObject json=jsonParser.makeHttpRequest(url_insertTo_outtrans,"POST",params);

【讨论】：

谢谢老兄！你拯救了我的一天。上帝祝福你。我会尽快接受你的答复。