Question

我想离开连接表来显示结果，我的查询如下：

$select= k_people,k_loans,k_loan_types CONCAT(customer.first_name,customer.last_name) as customer_name, 
                   CONCAT(agent.first_name,agent.last_name) as agent_name, 
                   CONCAT(approver.first_name,approver.last_name) as approver_name,
                   k_loan_types.name as k_loan_type FROM k_loans
       LEFT join k_people as customer', 'customer.person_id = k_loans.customer_id,

    LEFT join k_people as agent', 'agent.person_id = k_loans.loan_agent_id,
       LEFT join k_people as approver', 'approver.person_id = k_loans.loan_approved_by_id,
       LEFT join k_loan_types', 'k_loan_types.loan_type_id = k_loans.loan_type_id;
      $result= mysqli_query($con,select($select));

但它会引发以下错误：解析错误：语法错误，意外＆＃39;，＆＃39;在/ home / xXx /

任何人都可以帮助我理解我做错了，我的MYSQL查询知识非常有限，并试图通过doin来学习。

更新：在评论中应用建议后，查询如下所示：

$select=SELECT k_people,k_loans,k_loan_types CONCAT(customer.first_name,customer.last_name) as customer_name, 
                   CONCAT(agent.first_name,agent.last_name) as agent_name, 
                   CONCAT(approver.first_name,approver.last_name) as approver_name,
                   k_loan_types.name as k_loan_type FROM k_loans
       LEFT join k_people as customer ON customer.person_id = k_loans.customer_id,
       LEFT join k_people as agent ON agent.person_id = k_loans.loan_agent_id,
       LEFT join k_people as approver ON approver.person_id = k_loans.loan_approved_by_id,
       LEFT join k_loan_types ON k_loan_types.loan_type_id = k_loans.loan_type_id;
      $result= mysqli_query($con,$select);

错误更改为：解析错误：语法错误，意外＆＃39; kpos_people＆＃39; （T_STRING）

Answer 1

请删除额外的＆＃39;在您的字段名称后的查询中。它们不是必需的

select= k_people,k_loans,k_loan_types CONCAT(customer.first_name,customer.last_name) as customer_name, 
                   CONCAT(agent.first_name,agent.last_name) as agent_name, 
                   CONCAT(approver.first_name,approver.last_name) as approver_name,
                   k_loan_types.name as k_loan_type FROM k_loans
       LEFT join k_people as customer, customer.person_id = k_loans.customer_id,

    LEFT join k_people as agent,agent.person_id = k_loans.loan_agent_id,
       LEFT join k_people as approver, approver.person_id = k_loans.loan_approved_by_id,
       LEFT join k_loan_types, k_loan_types.loan_type_id = k_loans.loan_type_id;
      $result= mysqli_query($con,select($select));

在mysql查询中获取解析错误

1 个答案: