Joomla查詢選擇左加入$ var

Question

我需要得到左加入$var！沒有變量的代碼工作：

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

我試過所有輸入變量的語法，但它不工作！

$var="d.test"; 

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

我試過"$test" {$test}但它不起作用。

Answer 1

您正在傳遞帶單引號的字符串文字，這是問題所在。 PHP不會在用單引號定義的字符串中插入變量。他們更改爲雙引號來解決：

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

或者將這些變量：

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

從Manual：

不像雙引號和定界符語法，變量和轉義序列對於特殊字符，當它們出現在單引號字符串中時不會被擴展。

Answer 2

您正在使用單引號。那麼它不會將$ test轉換爲propper值。使用雙引號或使用$ test。'作爲部門'

$test = 'd.test'; 
print '$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