使用Like时mysql除以零错误

Question

我正在尝试使用PHP更新列数据，但我收到零错误的分组。

$find= "INSERT INTO closed_route( name, lat, latasal ) 
SELECT markers.name, markers.lat, user_locationtemp.lat AS some_name
FROM markers, user_locationtemp
WHERE markers.lat LIKE CONCAT( user_locationtemp.lat,  "%" )"; 
$result = mysql_query($find); 

Answer 1

问题在于您的报价。当您编写"%"时，您将结束以"INSERT开头的字符串。所以你在两个字符串上运行PHP %运算符，这是模数。由于"' )"不是数字，因此将其视为零，因此您将被除以零。

更改为字符串中的单引号：

$find= "INSERT INTO closed_route( name, lat, latasal ) 
SELECT markers.name, markers.lat, user_locationtemp.lat AS some_name
FROM markers, user_locationtemp
WHERE markers.lat LIKE CONCAT( user_locationtemp.lat,  '%' )"; 

Answer 2

在%附近使用单引号

$find= "INSERT INTO closed_route( name, lat, latasal ) 
SELECT markers.name, markers.lat, user_locationtemp.lat AS some_name
FROM markers, user_locationtemp
WHERE markers.lat LIKE CONCAT( user_locationtemp.lat,  '%' )"; 
$result = mysql_query($find);