0

我有更新 Mysql Db 的 php 页面。我不明白为什么我的以下 php 代码说“无法更新标记。未选择数据库”。奇怪的！！你能告诉我为什么它显示错误信息。

谢谢。

php代码：

<?php

// database settings 
$db_username = 'root';
$db_password = '';
$db_name = 'parkool';
$db_host = 'localhost';

//mysqli
$mysqli = new mysqli($db_host, $db_username, $db_password, $db_name);

if (mysqli_connect_errno()) 
{
header('HTTP/1.1 500 Error: Could not connect to db!'); 
exit();
}

################ Save & delete markers #################
if($_POST) //run only if there's a post data
{
//make sure request is comming from Ajax
$xhr = $_SERVER['HTTP_X_REQUESTED_WITH'] == 'XMLHttpRequest'; 
if (!$xhr){ 
    header('HTTP/1.1 500 Error: Request must come from Ajax!'); 
    exit(); 
}

// get marker position and split it for database
$mLatLang   = explode(',',$_POST["latlang"]);
$mLat       = filter_var($mLatLang[0], FILTER_VALIDATE_FLOAT);
$mLng       = filter_var($mLatLang[1], FILTER_VALIDATE_FLOAT);




$mName      = filter_var($_POST["name"], FILTER_SANITIZE_STRING);
$mAddress   = filter_var($_POST["address"], FILTER_SANITIZE_STRING);
$mId        = filter_var($_POST["id"], FILTER_SANITIZE_STRING);

/*$result = mysql_query("SELECT id FROM test.markers WHERE test.markers.lat=$mLat  
AND test.markers.lng=$mLng");
if (!$result) {
echo 'Could not run query: ' . mysql_error();
exit;
}
$row = mysql_fetch_row($result);
$id=$row[0];*/
//$output = '<h1 class="marker-heading">'.$mId.'</h1><p>'.$mAddress.'</p>';
//exit($output);
//Update Marker
if(isset($_POST["update"]) && $_POST["update"]==true)
{

    $results = mysql_query("UPDATE parkings SET latitude = '$mLat', longitude 
= '$mLng' WHERE locId = '94' ");
    if (!$results) {  
      //header('HTTP/1.1 500 Error: Could not Update Markers! $mId'); 
      echo "coudld not update marker."  . mysql_error();
      exit();
    } 

    exit("Done!");
}

$output = '<h1 class="marker-heading">'.$mName.'</h1><p>'.$mAddress.'</p>';
exit($output);
}


############### Continue generating Map XML #################

//Create a new DOMDocument object
$dom = new DOMDocument("1.0");
$node = $dom->createElement("markers"); //Create new element node
$parnode = $dom->appendChild($node); //make the node show up 

// Select all the rows in the markers table
$results = $mysqli->query("SELECT * FROM parkings WHERE 1");
if (!$results) {    
header('HTTP/1.1 500 Error: Could not get markers!'); 
exit();
} 

//set document header to text/xml
header("Content-type: text/xml"); 

// Iterate through the rows, adding XML nodes for each
while($obj = $results->fetch_object())
{
 $node = $dom->createElement("marker");  
$newnode = $parnode->appendChild($node);   
$newnode->setAttribute("name",$obj->name);
$newnode->setAttribute("locId",$obj->locId);
$newnode->setAttribute("address", $obj->address);  
$newnode->setAttribute("latitude", $obj->latitude);  
$newnode->setAttribute("longitude", $obj->longitude);  
//$newnode->setAttribute("type", $obj->type);   
}

echo $dom->saveXML();
4

2 回答 2

1

您正在将 mysqli 与 mysql 库混合使用。看，在这里你连接到数据库，使用 MySQLi 库：

$mysqli = new mysqli($db_host, $db_username, $db_password, $db_name);

在这里你正在使用 mysql_* 函数：

$results = mysql_query("UPDATE parkings SET latitude = '$mLat', longitude 
= '$mLng' WHERE locId = '94' ");
    if (!$results) {  
      //header('HTTP/1.1 500 Error: Could not Update Markers! $mId'); 
      echo "coudld not update marker."  . mysql_error();
      exit();
    }

当然，你会得到一个错误。

请将所有内容转换为 MySQLi，它应该可以工作。

于 2013-11-04T03:51:45.197 回答
0

也许您正试图在 mysqli 连接上执行 mysql 查询

$mysqli = new mysqli($db_host, $db_username, $db_password, $db_name);

$results = mysql_query("UPDATE parkings SET latitude = '$mLat', longitude = '$mLng' WHERE locId = '94' ");
于 2013-11-04T03:50:32.787 回答