php - 1064 mySQL 错误 : AS salary ON resume. jobsalaryrange = salary.id ' 在第 6 行

Question

我收到这个错误:

1064 You have an error in your SQL syntax; AS salary ON resume.jobsalaryrange = salary.id ' at line 6 SQL=SELECT resume.*, cat.cat_title, jobtype.title AS jobtypetitle , salary.rangestart, salary.rangeend , currency.symbol FROM_js_job_resumeAS resume JOIN_js_job_jobtypesAS jobtype ON resume.jobtype = jobtype.id LEFT JOIN_js_job_currenciesAS currency ON currency.id = resume.currencyid AND currency.id = LEFT JOIN_js_job_salaryrangeAS salary ON resume.jobsalaryrange = salary.id ,_js_job_categoriesAS cat WHERE resume.job_category = cat.id AND resume.status = 1 AND resume.searchable = 1 AND resume.nationality = 'KE' AND resume.iamavailable = 20 AND resume.jobtype = 2

来自这个查询:

    $db->setQuery($query);
    $total = $db->loadResult();

    if ( $total <= $limitstart ) $limitstart = 0;
    $query = "SELECT resume.*, cat.cat_title, jobtype.title AS jobtypetitle
            , salary.rangestart, salary.rangeend , currency.symbol
            FROM `#__js_job_resume` AS resume
            JOIN `#__js_job_jobtypes` AS jobtype ON resume.jobtype = jobtype.id
            LEFT JOIN `#__js_job_currencies` AS currency ON currency.id = resume.currencyid     AND currency.id =  " .$currency."       
            LEFT JOIN `#__js_job_salaryrange` AS salary ON resume.jobsalaryrange = salary.id
            , `#__js_job_categories` AS cat ";
        $query .= "WHERE resume.job_category = cat.id AND resume.status = 1 AND resume.searchable = 1";
    $query .= $wherequery;

有人知道吗？ 这对我来说很难!

Answer 1

它说

AND currency.id = LEFT JOIN_js_job_salaryrange AS 

所以好像没有设置变量$currency。

如果您在查询之前添加:$currency = 1; 只是为了测试查询，它可能会起作用。您只需找出货币 ID 应该是什么。