使用动态键和值在php中创建关联数组

Question

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.

表名是monthly_salary与名为两列month和salary分别.

我在里面得到数据:

$sql = mysql_query('SELECT * FROM monthly_salary');
$sql2 = mysql_query('SELECT * FROM monthly_salary');

然后分配,然后连接起来收集到的数据$mon和$sal:

$mon = "";
$sal = "";
while($row = mysql_fetch_array($sql)){
    $mon .= $row['month'].", ";
}
while($row = mysql_fetch_array($sql2)){
    $sal .= $row['salary'].", ";
}

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));

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;
}

那么如何修复它并使用正确的输出Jan => 3500？

Answer 1

你这个问题过于复杂了.这可以简单地完成,只需更少的循环.

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;
}

Answer 2

你为什么不这样做:

$associativeArray = array();
while($row = mysql_fetch_array($sql)){
    $associativeArray[$row['month']] = $row['salary'];
}