LIKE SQL 语法错误

Question

我正在使用 nodeJs 和 mysql 包。

我想使用带有变量的 LIKE sql 语句。

这是源代码：

var likemobile = '%'+mobile;
var query = "SELECT vtiger_contactaddress.contactaddressid as 'leadid', 
                    vtiger_contactaddress.mobile,
                    vtiger_contactaddress.phone 
             FROM `vtiger_contactaddress` 
             INNER JOIN `vtiger_crmentity` 
                 ON vtiger_crmentity.crmid=vtiger_contactaddress.contactaddressid AND 
                    vtiger_crmentity.deleted=0 AND 
                    vtiger_contactaddress.mobile LIKE "+likemobile+" OR 
                    vtiger_contactaddress.phone LIKE "+likemobile;

这是返回的错误：

Error: ER_PARSE_ERROR: You have an error in your SQL syntax; check the manual th
at corresponds to your MySQL server version for the right syntax to use near '%8
8436500 OR vtiger_leadaddress.phone LIKE %88436500' at line 1

Answer 1

您需要用单引号将类似的模式括起来。

另外两个建议：

• 仅对字符串和日期常量（而不是列别名）使用单引号。
• 使用参数化查询，因此您不会将用户输入放入查询字符串中。

Answer 2

如果您确实在使用node-mysql，则应按照the documentation 中的说明运行查询。假设您已经有一个connection 对象，使用绑定变量进行查询就变得简单了：

connection.query({
  sql : "SELECT vtiger_contactaddress.contactaddressid as leadid, "
                   + " vtiger_contactaddress.mobile, "
                   + " vtiger_contactaddress.phone "
             + "FROM `vtiger_contactaddress` "
             + "INNER JOIN `vtiger_crmentity` "
             + "  ON vtiger_crmentity.crmid=vtiger_contactaddress.contactaddressid"
             + " AND vtiger_crmentity.deleted=0 AND "
             + "       (vtiger_contactaddress.mobile LIKE concat('%', ?) OR "
             + "       vtiger_contactaddress.phone LIKE concat('%', ?))",
  values: [mobile, mobile]
}, function (error, results, fields) {
  // error will be an Error if one occurred during the query
  // results will contain the results of the query
  // fields will contain information about the returned results fields (if any)
});

注意事项：

• 使用绑定变量，防止SQL Injection
• concat 函数用于在（已清理的）输入前加上通配符 (%)
• ORed 在一起的两个查询条件现在用括号分隔 vtiger_crmentity.deleted=0，这可能是您想要的
• 需要在回调函数中编写结果集处理代码，通过results和fields变量访问数据

Answer 3

@Gordon Linoff 是正确的。有点像这样：

var likemobile = mobile;
var query = "SELECT vtiger_contactaddress.contactaddressid as 'leadid', vtiger_contactaddress.mobile, vtiger_contactaddress.phone FROM `vtiger_contactaddress` INNER JOIN `vtiger_crmentity` ON vtiger_crmentity.crmid=vtiger_contactaddress.contactaddressid AND vtiger_crmentity.deleted=0 AND vtiger_contactaddress.mobile LIKE '%"+likemobile+"%' OR vtiger_contactaddress.phone LIKE '%"+likemobile+"%'";

Answer 4

... AND vtiger_contactaddress.mobile LIKE '%" + likemobile + "%' OR vtiger_contactaddress.phone LIKE %" + likemobile + "%";

Answer 5

try this..

         var likemobile = mobile;

var query = "SELECT vtiger_contactaddress.contactaddressid as 'leadid', 
                vtiger_contactaddress.mobile,
                vtiger_contactaddress.phone 
         FROM `vtiger_contactaddress` 
         INNER JOIN `vtiger_crmentity` 
             ON vtiger_crmentity.crmid=vtiger_contactaddress.contactaddressid AND 
                vtiger_crmentity.deleted=0 AND 
                vtiger_contactaddress.mobile LIKE '% "+likemobile+"%'  OR 
                vtiger_contactaddress.phone LIKE '%"+likemobile +"%'";