我收到此错误：
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_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 = LEFT JOIN _js_job_salaryrange AS salary ON resume.jobsalaryrange = salary.id , _js_job_categories AS 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
从此查询：
$db->setQuery($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 .= $wherequery;
有人有线索吗？
对我来说这是一个艰难的过程！
最佳答案
它说
AND currency.id = LEFT JOIN_js_job_salaryrange AS
因此，似乎未设置变量$ currency。
如果仅在查询之前添加：
$currency = 1;仅用于测试查询，则它可能会起作用。您只需要找出货币ID应该是什么即可。关于php - 1064 mySQL错误:在第6行中AS工资在resume.jobsalaryrange =工资.id'，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/22105136/