Question

我在数据库上写了一个查询，结果以这种形式给出

Apples-100
Grapes-200
Oranges-50
Bananas-30

现在我需要将数据存储在像这样的数组中

[['Apples',100],['Grapes',200],['Oranges',50],['Bananas',30]]

我怎么能实现这一点我试过这个

while($row = mysql_fetch_assoc($getdatagraoh)){ 
    $plotgraphs[] = array(
        $row['Compliancestatus'],
        $row['value']
    );
}
header('ContentType: application/json; charset=utf-8');
$data= json_encode($plotgraphs);

我无法实现这一点。请在这方面帮助我。

Answer 1

首先不要使用mysql_ *作为其删除，而是使用mysqli_ *或PDO。

在$plotgraphs中存储值后，回到您的问题，然后像这样爆炸，您将获得所需的结果

foreach ( $array as $k => $v ) {
    $seprated = explode ( '-', $v );

    //echo '<pre>';
    //print_r ( $seprated );
    echo json_encode(( $seprated ));
}

输出

["Apples","100"]["Grapes","200"]["Oranges","50"]["Bananas","30"]

Answer 2

while($row = mysql_fetch_assoc($getdatagraoh)){ 
  $plotgraphs[] =$row;
}

旁注：Why shouldn't I use mysql_* functions in PHP?

Answer 3

试试这个：

<?php
$plotgraphs = array();

while($row = mysql_fetch_assoc($getdatagraoh)){ 
    $plotgraphs[] = $row;
}
?>

此$row数组已分配给$plotgraphs。

Answer 4

var $plotgraphs = [];  
  while( $row = mysql_fetch_assoc( $getdatagraoh)){
        $plotgraphs[] = $row; // Inside while loop
        //or
        $plotgraphs[ $row['id']] = $row;
    }

Answer 5

试试这样：既然您有更改问题，可以试试这个：

$arrayName = array('Apples-100','Grapes-200','Oranges-50','Bananas-30');



foreach($arrayName as $value)
{
     $plotgraphs = explode("-",$value);
   echo  json_encode( $plotgraphs);
}

输出：[“苹果”，“100”] [“葡萄”，“200”] [“橘子”，“50”] [“香蕉”，“30”]

OR

foreach($arrayName as $value)
{
     $plotgraphs[] = explode("-",$value);

}
echo json_encode( $plotgraphs);

如何在PHP中将查询结果导入数组

5 个答案: