Question

 $sql= 'select employee.id AS ID,CONCAT(employee.fname," ",employee.lname) AS Name, IF(loan.status= "unpaid",loan.installment,0) AS Loan where employee.user = "$name"';

这不会显示表格的结果。因为此查询显示不匹配。我想把＆＃34; $ name＆＃34;双引号是错误。

   $sql= "select employee.id AS ID,CONCAT(employee.fname," ",employee.lname) AS Name, IF(loan.status= 'unpaid',loan.installment,0) AS Loan where employee.user = '$name'"

在这种情况下，将“未付”＆＃39;单引号中没有显示此列的结果

我试过了 employee.user =＆＃39;＆＃39;＆＃39; $名称＆＃39;＆＃39;＆＃39; employee.user =＆＃34;＆＃39; $ name＆＃39;＆＃34; 这些工作都没有..请帮忙

Answer 1

引号序列似乎正确但是你错过了从句子

$sql= "select employee.id AS ID,CONCAT(employee.fname, ' ' ,employee.lname) AS Name, 
    IF(loan.status = 'unpaid',loan.installment,0) AS Loan 
    FROM   employee
    INNER JOIN loan on loan.your_key1 = employee.your_key2
    where employee.user = '$name'";

您必须为your_key1指定正确的列名，为JOIN指定your_key2

php Mysql引号错误

1 个答案: