Question

我在PHP中有以下代码：

$stmt2 = $dbh->prepare("select GROUP_CONCAT( cohort_id SEPARATOR ',') as cohort_id from ( select distinct cohort_id from table_cohorts) as m"); 
$stmt2->execute(); 
$row2 = $stmt2->fetch();
$cohorts_allowed = explode(",",$row2["cohort_id"]);
$value = array ('amount', $cohorts_allowed );

$cohorts_allowed给了我类似于＆＃34; database_percent，national_percent＆＃34;的内容。它是从我的数据库中所有可能的cohort_id生成的。

我需要做的是获取所有这些并添加额外的价值＆＃39;金额＆＃39; （不在我的数据库中）进入数组。

我该怎么做？您可以看到我尝试在上面的代码的最后一行执行此操作，但显然这不起作用。

Answer 1

$cohorts_allowed = explode(",",$row2["cohort_id"]);
$cohorts_allowed['amount'] = 'amount' ;

或

$cohorts_allowed = explode(",",$row2["cohort_id"]);
$cohorts_allowed[] = 'amount' ;

这是如何运作的：

<pre>
<?php

$row2["cohort_id"] = "database_percent, national_percent";
$cohorts_allowed = explode(",",$row2["cohort_id"]);

print_r($cohorts_allowed);

/* output
Array
(
    [0] => database_percent
    [1] =>  national_percent

)
* The last index is 1
*/

$cohorts_allowed[] = 'amount' ;

print_r($cohorts_allowed);

/* output
Array
(
    [0] => database_percent
    [1] =>  national_percent
    [2] => amount
)
* The last index is 2 (after 1) and have the value amount.
* 
*/

?>
</pre>

您可以阅读：

http://php.net/manual/en/language.types.array.php

为从PHP中的变量构建的数组添加其他值

1 个答案: