php - Joomla 查询用 $var 选择左连接

Question

我需要用一个左连接$var！没有变量，代码可以工作：

$query->select('d.value AS department');
$query->join('LEFT', '#__jea_departments AS d ON d.id = p.department_id');

我尝试了输入变量的所有语法，但它不起作用！

$var="d.test";

$query->select('**$test** AS department');
$query->join('LEFT', '#__jea_departments AS d ON d.id = p.department_id');

我试过"$test" {$test}了，但它不起作用。

Answer 1

您正在使用单引号。然后它不会将 $test 转换为 propper 值。使用双引号或使用 $test 。作为部门'

$test = 'd.test';
print '$test as department';  //will print: $test as department
print "$test as department";  //will print: d.test as department
print $test. ' as department';  //will print: d.test as department

Answer 2

您正在使用单引号传递字符串文字，这就是问题所在。PHP 不会在用单引号定义的字符串中插入变量。将它们更改为双引号以解决：

$query->select("$test AS department");

或者连接变量：

$query->select($test . ' AS department');

从手册：

与双引号和heredoc 语法不同，特殊字符的变量和转义序列在出现在单引号字符串中时不会被扩展。