Php MySQL Select查询参数之间的多个变量，变量不起作用

我正在开发一个应用程序，对于给定的位置（坐标lat/long对），它显示位于给定起始位置的给定半径内的所有实体（从数据库中提取）

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

我已经测试过了，我用来查找最近位置的逻辑非常有效。然而，在与MySQL表通信的PHP脚本中，我遇到了一个问题

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

这是我正在尝试的查询，没有从表中得到任何结果：

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

但是，通过回显$lat1、$lat2、$lon1和$lon2，我知道它们包含有效值（这应该会产生查询结果）（仅供参考：这4个变量表示给定位置的半径平方；纬度和经度是表中的列。）

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

但是，这是可行的（只是在值中硬编码，不使用$lon2变量引用它：

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

$sql = "SELECT * FROM cashbackengine_merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND **-90.2745** AND category='$merchCategory'";            

正如这一点：

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

$sql = "SELECT * FROM cashbackengine_merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN **-90.64484** AND '$lon2' AND category='$merchCategory'";           

但纬度一点问题都没有。我强烈感觉这与语法有关，而且可能非常简单。非常感谢任何帮助！

试试这个

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN $lon1 AND $lon2 AND category='$merchCategory'";

我不确定，但我认为这是因为BETWEEN需要两个数字作为您的案例，您将其作为字符串传递

'$lon'

。当您硬编码一个值时，另一个值将自动转换为数字值

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";           

---

旁注：查看准备好的语句（best）或转义字符串，因为如果用户正在输入lat/lon值，这将打开sql注入。

变量应该绑定到准备好的语句，而不是注入到查询中。此页面上共享潜在的不安全性。

$sql = "SELECT * FROM merchants WHERE latitude BETWEEN '$lat1' AND '$lat2' AND longitude BETWEEN '$lon1' AND -90.2745 AND category='$merchCategory'";