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使用PHP更新MySQL数据库

[英]Updating MySQL database with PHP

基本上,我试图使用PHP更新MySQL数据库,并使用HTML表单对其进行测试。

我打算在一个Android应用程序中使用它,以便从中获取值,但是目前我仅使用HTML表单进行测试以测试PHP代码。 当我使用HTML表单进行测试时,相应的数据当前未更新。

我的代码导致了什么问题?

PHP代码:

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

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

// check for required fields
if (isset($_POST['PlayerID']) && isset($_POST['Score']) && isset($_POST['LastHolePlayed'])&&     
isset($_POST['Overall'])) {

$playerid = $_POST['PlayerID'];
$score = $_POST['Score'];
$lastholeplayed = $_POST['LastHolePlayed'];
$overall = $_POST['Overall'];

// include db connect class
require('db_connection.php');



// mysql inserting a new row
$result = mysql_query("UPDATE `week1` SET Score = `$score`, LastHolePlayed = `$lastholeplayed`, 

Overall` = $overall` WHERE PlayerID = `$playerid`");


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

    // echoing JSON response
    echo json_encode($response);
} else {
    // failed to insert row
    $response["success"] = 0;
    $response["message"] = "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);
}

html代码:

<form action="http://localhost/realdeal/updateplayer.php" method="POST">
PlayerID <input type="text" id='PlayerID' name='PlayerID'><br/><br/>
Score <input type="text" id='Score' name='Score'><br/><br/>
LastHolePlayed <input type="text" id='LastHolePlayed' name='LastHolePlayed'><br/><br/>
Overall <input type="text" id='Overall' name='Overall'><br/><br/>

    &nbsp;  <input type="submit" value="submit">

</form>

将查询更改为:

$result = mysql_query("UPDATE `week1` SET `Score` = '$score', `LastHolePlayed` = '$lastholeplayed', `Overall` = '$overall' WHERE `PlayerID` = '$playerid'");

您的查询定界符需要更正:

$result = mysql_query("UPDATE `week1` SET Score = '$score', `LastHolePlayed` = '$lastholeplayed', `Overall` = '$overall' WHERE `PlayerID` = '$playerid'");

请注意,列名和表名周围的反引号(`)和值周围的单引号(')。

另外,调试查询时,请始终检查MySQL错误:

$result mysql_query(...) or die("Query failed: " . mysql_error() );

最后,您应该知道查询使您容易受到SQL注入攻击的影响。 在将输入数据包括在查询中之前,请始终对其进行清理。

您的sql语句错误。 您可以如上所述编写,也可以直接编写不带撇号的语句-$ result = mysql_query(“ UPDATE week1 SET Score = $ score,LastHolePlayed = $ lastholeplayed,总体= $ overall WHERE PlayerID = $ playerid”);

此外,您能否解释一下“未更新适当的数据”是什么意思。 如果您给出/陈述所得到的错误,将会更加清楚。

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