我有两个索引为1到10和t的数组，当我必须根据索引进行合并时，所以只需要添加一些员工Salary。我已经使用array_merge（）完成了，但是输出没有合并我无法获得正确的输出。输出仅是employeeSalary Array。为什么？我已经尝试过了，但是无法正常工作。
$employee = array
(
0=>
array("employee_id"=>1, "firstName"=>"Zahir", "lastName"=>"Alam", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
,"Head"=>
array("Id"=>3 , "Name"=>"Sourasis Roy")
)
,
1=>
array("employee_id"=>2, "firstName"=>"Amith", "lastName"=>"Manniken", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
,"Head"=>
array("Id"=>3 , "Name"=>"Sourasis Roy")
)
,
2=>
array("employee_id"=>3, "firstName"=>"Sourasis", "lastName"=>"Roy", "Age"=>28, "Company"=>"Switchme", "Role"=>"CTO")
,
3=>
array("employee_id"=>4, "firstName"=>"Aditya", "lastName"=>"Mishra", "Age"=>29, "Company"=>"Switchme", "Department"=>"Tech", "Role"=>"CEO")
,
4=>
array("employee_id"=>5, "firstName"=>"Priti", "lastName"=>"Lata", "Age"=>24, "Company"=>"Switchme", "Role"=>"HR")
,
5=>
array("employee_id"=>6, "firstName"=>"Sumita", "lastName"=>"Nath", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
6=>
array("employee_id"=>7, "firstName"=>"Tarini", "lastName"=>"Khanna", "Age"=>22, "Company"=>"Switchme", "Role"=>"Content Writer")
,
7=>
array("employee_id"=>8, "firstName"=>"Abhisek", "lastName"=>"Soni", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm","Head"=>array("Id"=>5 , "Name"=>"Sumita Nath")
)
,
8=>
array("employee_id"=>9, "firstName"=>"Ankit", "lastName"=>"Pump", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm"
,"Head"=>
array("Id"=>5 , "Name"=>"Sumita Nath")
)
,
9=>
array("employee_id"=>10, "firstName"=>"Pogo", "lastName"=>"Laal", "Age"=>23, "Company"=>"Switchme", "Role"=>"Designer")
,
10=>
array("employee_id"=>11, "firstName"=>"Sabina", "lastName"=>"Sekh", "Age"=>28, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
11=>
array("employee_id"=>12, "firstName"=>"Sanjay", "lastName"=>"Poudal", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm"
,"Head"=>
array("Id"=>10 , "Name"=>"Sabina Sekh")
)
,
);
$employee_salary = array
(
7=>
array("employee_id"=>7, "salary"=>"55,000"
)
,
2=>
array("employee_id"=>2, "salary"=>"60,000"
)
,
9=>
array("employee_id"=>9, "salary"=>"50,000"
)
,
10=>
array("employee_id"=>10, "salary"=>"30,000"
)
,
);
$ar= array();
for($j=0;$j<count($employee);$j++)
{
error_reporting(0);
if($employee[$j]=$employee_salary[$j])
{
$ar=$employee[$j]+$employee_salary[$j];
print_r($ar);
echo "<br>";
}
error_reporting(0);
}
答案 0 :(得分：2)
正如克里斯·怀特（chris white）所说，您忘记再添加一个=作为比较。并使用 isset（）检查employee_salary的第二个数组中是否设置了数组索引。例如 if（$ employee [$ j] == isset（$ employee_salary [$ j]））
$employee = array
(
0=>
array("employee_id"=>1, "firstName"=>"Zahir", "lastName"=>"Alam", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
,"Head"=>
array("Id"=>3 , "Name"=>"Sourasis Roy")
)
,
1=>
array("employee_id"=>2, "firstName"=>"Amith", "lastName"=>"Manniken", "Age"=>25, "Company"=>"Switchme", "Role"=>"Developer", "Department"=>"Tech"
,"Head"=>
array("Id"=>3 , "Name"=>"Sourasis Roy")
)
,
2=>
array("employee_id"=>3, "firstName"=>"Sourasis", "lastName"=>"Roy", "Age"=>28, "Company"=>"Switchme", "Role"=>"CTO")
,
3=>
array("employee_id"=>4, "firstName"=>"Aditya", "lastName"=>"Mishra", "Age"=>29, "Company"=>"Switchme", "Department"=>"Tech", "Role"=>"CEO")
,
4=>
array("employee_id"=>5, "firstName"=>"Priti", "lastName"=>"Lata", "Age"=>24, "Company"=>"Switchme", "Role"=>"HR")
,
5=>
array("employee_id"=>6, "firstName"=>"Sumita", "lastName"=>"Nath", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
6=>
array("employee_id"=>7, "firstName"=>"Tarini", "lastName"=>"Khanna", "Age"=>22, "Company"=>"Switchme", "Role"=>"Content Writer")
,
7=>
array("employee_id"=>8, "firstName"=>"Abhisek", "lastName"=>"Soni", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm","Head"=>array("Id"=>5 , "Name"=>"Sumita Nath")
)
,
8=>
array("employee_id"=>9, "firstName"=>"Ankit", "lastName"=>"Pump", "Age"=>23, "Company"=>"Switchme", "Role"=>"HLA", "Department"=>"Crm"
,"Head"=>
array("Id"=>5 , "Name"=>"Sumita Nath")
)
,
9=>
array("employee_id"=>10, "firstName"=>"Pogo", "lastName"=>"Laal", "Age"=>23, "Company"=>"Switchme", "Role"=>"Designer")
,
10=>
array("employee_id"=>11, "firstName"=>"Sabina", "lastName"=>"Sekh", "Age"=>28, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm")
,
11=>
array("employee_id"=>12, "firstName"=>"Sanjay", "lastName"=>"Poudal", "Age"=>24, "Company"=>"Switchme", "Role"=>"HLA Head", "Department"=>"Crm"
,"Head"=>
array("Id"=>10 , "Name"=>"Sabina Sekh")
)
,
);
$employee_salary = array
(
7=>
array("employee_id"=>7, "salary"=>"55,000"
)
,
2=>
array("employee_id"=>2, "salary"=>"60,000"
)
,
9=>
array("employee_id"=>9, "salary"=>"50,000"
)
,
10=>
array("employee_id"=>10, "salary"=>"30,000"
)
,
);
$ar= array();
for($j=0;$j<count($employee);$j++)
{
error_reporting(0);
if($employee[$j]==isset($employee_salary[$j]))
{
$ar=$employee[$j]+$employee_salary[$j];
print_r($ar);
echo "<br>";
}
error_reporting(0);
}
答案 1 :(得分：0)
使用问题中给出的数组UIImageView?和class Thing {
var iv: UIImageView? = UIImageView()
func test() {
let im = UIImage()
let kp = \UIImageView.image
self.iv?[keyPath: kp] = im
}
}
的值，尝试以下解决方案。
class Thing {
var iv: UIImageView! = UIImageView()
func test() {
let im = UIImage()
let kp = \UIImageView.image
self.iv![keyPath: kp] = im
}
}
答案 2 :(得分：0)
我认为也将有一个简单的解决方案。因为您已经使用id键创建了employee_salary数组。所以会是这样。
Foreach($employee as $key=> $value)
{
$employee[$key]['salary'] = $employee_salary[$value['employee_id']]['salary'];
}
Print_r($employee);die;
我认为这会帮助您。谢谢
答案 3 :(得分：0)
那应该做
$ar= array();
for($j=0;$j<count($employee);$j++)
{
$ar[$j]=$employee[$j];
if(isset($employee_salary[$employee[$j]["employee_id"]]) && $employee_salary[$employee[$j]["employee_id"]]['employee_id'] == $employee[$j]["employee_id"] ){
$ar[$j]=array_merge($employee[$j],$employee_salary[$employee[$j]["employee_id"]]);
}
}
请注意阵列键。