Question

我尝试存储使用javascript获取的浏览器地理位置，并通过$.ajax通过php发布到mysql表中。我收到发布的数据并以递归数组转换，因此我只能获得纬度和经度数据，但我得到两个警告（我将对代码进行评论）：

1->Warning: mysqli_real_escape_string() expects parameter 2 to be string, object given in. 2->Warning: mysqli_error() expects exactly 1 parameter, 0 given in

这是我的代码：

地理定位并发送数据：

if (window.navigator.geolocation) {
    var failure, success;
    success = function (position) {
        console.log(position);
        var stringData = JSON.stringify(position, null);
        $.ajax({
            type: "POST",
            url: "GL_.php",
            data: {
                data: stringData
            }
        });
    };
    failure = function (message) {
        alert('Cannot retrieve location!');
    };
    navigator.geolocation.getCurrentPosition(success, failure, {
        maximumAge: Infinity,
        timeout: 5000
    });
}

...接收数据 - ＆gt; ...

<? php
$hostname_connection = "p:localhost";
$database_connection = "s_c"
$username_connection = "root";$password_connection = "";
$cs_connection = mysqli_connect($hostname_connection, $username_connection, $password_connection, $database_connection) or trigger_error(mysqli_error(), E_USER_ERROR); mysqli_set_charset($cs_connection, 'utf8');

function mysqli_result($res, $row, $field = 0) {
    $res - > data_seek($row);
    $datarow = $res - > fetch_array();
    return $datarow[$field];
}
if (!function_exists("GetSQLValueString")) {
    function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") {
        if (PHP_VERSION < 6) {
            $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
        }
        global $cs_connection;
        $theValue = > function_exists("mysqli_real_escape_string") ? mysqli_real_escape_string($cs_connection, $theValue) : mysqli_escape_string($theValue); //FIRST WARNING  
        switch ($theType) {
            case "text":
                $theValue = ($theValue != "") ? "'".$theValue."'" : "NULL";
                break;
            case "long":
            case "int":
                $theValue = ($theValue != "") ? intval($theValue) : "NULL";
                break;
            case "double":
                $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
                break;
            case "date":
                $theValue = ($theValue != "") ? "'".$theValue."'" : "NULL";
                break;
            case "defined":
                $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
                break;
        }
        return $theValue;
    }
}
if (isset($_POST['data'])) {
    $dataString = $_POST['data'];
}
function geoCodeUser($dataString) {
    global $database_connection;
    global $cs_connection;
    $position = json_decode($dataString, true);
    $lat = $position['coords']['latitude'];
    $lng = $position['coords']['longitude'];
    if ($dataString !== NULL) {
        $insertLatLng = sprintf("INSERT INTO usergeoloc (lat,long) VALUES (%s, %s)", GetSQLValueString($cs_connection, $lat, "text"), GetSQLValueString($cs_connection, $lng, "text"));
        $Result1 = mysqli_query($cs_connection, $insertLatLng) or die(mysqli_error($cs_connection)); //SECOND WARNING
    } else {
        echo "NO CONTENT";
    }
}
geoCodeUser($dataString);
?>

变量$ lat和$ lng每个都填充相应的值，但正如我之前提到的，错误出现了。任何人都可以解释这里的错误吗？

Answer 1

对于第一个错误，您的问题是您调用GetSQLValueString方法错误，您将其定义为

GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "")

但是用这些参数调用它

GetSQLValueString($cs_connection, $lat, "text"),
GetSQLValueString($cs_connection, $lng, "text")

所以$ theValue被设置为一个对象（mysqli链接）

至于mysqli_error错误，你没有传递必需的参数

http://php.net/manual/en/mysqli.error.php

程序风格

字符串mysqli_error（mysqli $ link）

你有：

die(mysqli_error())

应该是

die(mysqli_error($cs_connection))

Answer 2

抓住这样：

if($cs_connection){
  $theValue = mysqli_real_escape_string($cs_connection, $theValue);
} else {
  // No DB Connection, so no way reason to escape
}

用php将浏览器地理位置插入mysql表？

2 个答案: