MySQL查询不在MySQLi中工作，但在MySQL中工作

我正在为iOS应用程序编写一个API，并且需要在地理坐标附近获取位置。MySQL查询不在MySQLi中工作，但在MySQL中工作

我在mysqli中试过这个查询，它没有返回任何结果。当我在一个常规的mysql_query中完成时，它完美地工作。我也在phpmyadmin中测试了它的成功完成的查询。

SELECT id, name, address, city, state, longitude, latitude, ($miles * acos(cos(radians($latitude)) * cos(radians(latitude)) * cos(radians(longitude) - radians($longitude)) + sin(radians($latitude)) * sin(radians(latitude)))) AS distance 
     FROM locations HAVING distance < $distance ORDER BY distance LIMIT 0, 20

这与变量查询填写：

SELECT id, name, address, city, state, longitude, latitude, (3959 * acos(cos(radians(40.735767)) * cos(radians(latitude)) * cos(radians(longitude) - radians(-73.991806)) + sin(radians(40.735767)) * sin(radians(latitude)))) AS distance 
     FROM locations HAVING distance < 25 ORDER BY distance LIMIT 0, 20

有我丢失的东西，为什么它不会在mysqli工作？

这是工作MySQL查询代码

mysql_connect('localhost', 'test', 'test') or die(mysql_error()); 
mysql_select_db('testdb') or die(mysql_error()); 
$locations = array(); 
$miles = 3959; 
$distance = 25; 
$latitude = "40.735767"; 
$longitude = "-73.991806"; 

$data = mysql_query("SELECT id, name, address, city, state, longitude, latitude, ($miles * acos(cos(radians($latitude)) * cos(radians(latitude)) * cos(radians(longitude) - radians($longitude)) + sin(radians($latitude)) * sin(radians(latitude)))) AS distance 
     FROM locations HAVING distance < $distance ORDER BY distance LIMIT 0, 20") 
    or die(mysql_error()); 

while ($row = mysql_fetch_assoc($data)) { 
    array_push($locations, $row); 
}

我使用MysqliDb类https://github.com/ajillion/PHP-MySQLi-Database-Class所以应该工作如下

 $row = $db->rawQuery("SELECT id, name, address, city, state, longitude, latitude, (? * acos(cos(radians(?)) * cos(radians(latitude)) * cos(radians(longitude) - radians(?)) + sin(radians(?)) * sin(radians(latitude)))) AS distance 
     FROM locations HAVING distance < ? ORDER BY distance LIMIT 0, 20", array($miles, $latitude, $longitude, $latitude, $distance)); 

     if (count($row) > 0){ 
      // if found, return JSON response 
      echo json_encode($row[0]); 
     }

即使当我用我发现mysqli的基本模板，它失败。

$locations = array(); 
$miles = 3959; 
$distance = 25; 
$latitude = "40.735767"; 
$longitude = "-73.991806"; 

// Connect to database 
$link = mysqli_connect('localhost','test','test','testdb'); 

// Check for Errors 
if(mysqli_connect_errno()){ 
    echo mysqli_connect_error(); 
} 

// Prepare Query 
$query = "SELECT id, name, address, city, state, longitude, latitude, ($miles * acos(cos(radians($latitude)) * cos(radians(latitude)) * cos(radians(longitude) - radians($longitude)) + sin(radians($latitude)) * sin(radians(latitude)))) AS distance 
     FROM locations HAVING distance < $distance ORDER BY distance LIMIT 0, 20"; 

// Escape Query 
$query = mysqli_real_escape_string($link,$query); 

// Perform Query 
if($result = mysqli_query($link,$query)){ 

// Cycle through results 
while($row = mysqli_fetch_object($result)){ 
    array_push($locations, $row); 
} 
// Free Result Set 
mysqli_free_result($result); 
} 

// Close Connection 
mysqli_close($link);

来源

2013-12-18 Wes

哪里调用mysql和mysqli函数？ – AbraCadaver

这个例子没有使'mysql'或'mysqli'调用的代码是没有用的。意思是，这些查询可能100％没有错，但也许你的PHP实现是关闭的。 – JakeGould

取决于您如何查询它。您只向我们展示一行代码。向我们展示您的完整代码。 –

有我丢失的东西，为什么它不会在mysqli的工作吗？

号
MySQL和mysqli的API的运行查询完全相同的方式。

寻找错别字。和其他类型的错误。

来源

2013-12-18 20:55:10

MySQL查询不在MySQLi中工作，但在MySQL中工作

回答

相关问题