I want to create an associative array in php with dynamic key and also a dynamic value from a particular mysql table.
我想在php中创建一个带有动态键的关联数组,以及来自特定mysql表的动态值。
The table name is monthly_salary with a two column named month and salary respectively.
表名是monthly_salary,分别有一个名为month和salary的列。
I get the data inside it:
我在里面得到数据:
$sql = mysql_query('SELECT * FROM monthly_salary');
$sql2 = mysql_query('SELECT * FROM monthly_salary');
Then assigned and concatenated the collected data to $mon
and $sal
:
然后将收集的数据分配并连接到$ mon和$ sal:
$mon = "";
$sal = "";
while($row = mysql_fetch_array($sql)){
$mon .= $row['month'].", ";
}
while($row = mysql_fetch_array($sql2)){
$sal .= $row['salary'].", ";
}
After that I've converted it to array and concatenate it until it became and associative array:
之后我将它转换为数组并连接它直到它成为关联数组:
$monArray = array(substr(trim($mon), 0, -1));
$salArray = array(substr(trim($sal), 0, -1));
$key = "";
$keyWithVal = "";
foreach($monArray as $k){
$key .= $k." => ";
}
foreach($salArray as $k){
$keyWithVal .= $key.$k.",";
}
$associativeArray = array(substr(trim($keyWithVal), 0, -1));
My Problem is that when I've echo it the result is always like this 3500=>Jan=>3500:
我的问题是,当我回应它时,结果总是像3500 => Jan => 3500:
foreach($associativeArray as $k => $id){
echo $k."=>".$id;
}
So how can I fix it and with the correct output Jan=>3500?
那么如何修复它并使用正确的输出Jan => 3500?
2 个解决方案
#1
23
You are way over-complicating this problem. This can be done simply, with fewer loops.
你这个问题过于复杂了。这可以简单地完成,具有更少的循环。
First, you only need to run the SQL once. Second, build the array in the 1st loop.
首先,您只需要运行一次SQL。其次,在第一个循环中构建数组。
$sql = mysql_query('SELECT * FROM monthly_salary');
$associativeArray = array();
while($row = mysql_fetch_array($sql)){
// Put the values into the array, no other variables needed
$associativeArray[$row['month']] = $row['salary'];
}
foreach($associativeArray as $k => $id){
echo $k."=>".$id;
}
#2
4
Why don't you just do:
你为什么不这样做:
$associativeArray = array();
while($row = mysql_fetch_array($sql)){
$associativeArray[$row['month']] = $row['salary'];
}
#1
23
You are way over-complicating this problem. This can be done simply, with fewer loops.
你这个问题过于复杂了。这可以简单地完成,具有更少的循环。
First, you only need to run the SQL once. Second, build the array in the 1st loop.
首先,您只需要运行一次SQL。其次,在第一个循环中构建数组。
$sql = mysql_query('SELECT * FROM monthly_salary');
$associativeArray = array();
while($row = mysql_fetch_array($sql)){
// Put the values into the array, no other variables needed
$associativeArray[$row['month']] = $row['salary'];
}
foreach($associativeArray as $k => $id){
echo $k."=>".$id;
}
#2
4
Why don't you just do:
你为什么不这样做:
$associativeArray = array();
while($row = mysql_fetch_array($sql)){
$associativeArray[$row['month']] = $row['salary'];
}