Question

我正在尝试以以下格式输出JSON：

[{
  "category": "cONTIME",
  "value": 155,
  "color": "#0f9b0f"
},{
  "category": "cLATE",
  "value": 244,
  "color": "#dd1818"
},{
  "category": "cTOTAL",
  "value": 399,
  "color": "#0575E6"
},{
  "category": "ONTIME",
  "value": 60,
  "color": "#0f9b0f"
},{
  "category": "LATE",
  "value": 58,
  "color": "#dd1818"
},{
  "category": "TOTAL",
  "value": 118,
  "color": "#0f9b0f"
}]

我正在执行的存储过程返回下表-这是SQL SERVER-我将不感兴趣的列涂成灰色

我尝试使用

$categories = [
    'cONTIME' => $row['cONTIME'],
    'cLATE' => [],
    'cTOTAL' => [],
];
$colors = [
    'ONTIME' => "#0f9b0f",
    'LATE' => "#dd1818",
    'TOTAL' => "#0575E6",
];

$json = array();

while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {
        $json['category'] = $categories;
        $json['value'] = $row;
        $json['color'] = $colors;
}
echo json_encode($json);

返回此：

{
   "category":{
      "cONTIME":null,
      "cLATE":[

      ],
      "cTOTAL":[

      ]
   },
   "value":{
      "cTOTAL":399,
      "cONTIME":155,
      "cLATE":244,
      "cPERCENTS":38,
      "TOTAL":118,
      "ONTIME":60,
      "LATE":58,
      "PERCENTS":50,
      "KPI":"UP"
   },
   "color":{
      "ONTIME":"#0f9b0f",
      "LATE":"#dd1818",
      "TOTAL":"#0575E6"
   }
}

我也尝试过这种方法：

$json = array();

while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {          
        $json[] = $row;
}
echo json_encode($json);

返回：

[
   {
      "cTOTAL":399,
      "cONTIME":155,
      "cLATE":244,
      "cPERCENTS":38,
      "TOTAL":118,
      "ONTIME":60,
      "LATE":58,
      "PERCENTS":50,
      "KPI":"UP"
   }
]

和上面一样，我尝试了许多其他方法和示例，但似乎无法实现我需要的格式。我完全了解PDO，Prepared Statements，Injection等！，我只需要首先格式化JSON，但在此先感谢您的建议

任何帮助将不胜感激-PHP Noob

Answer 1

类似的事情应该起作用：

$json = array();

while( $row = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) {          
    foreach ($row as $fieldName => $fieldValue) {
        if (in_array($fieldName, ['cTOTAL', 'cONTIME', 'cLATE', 'TOTAL' ,'ONTIME', 'LATE'])) {
            $json[] = [
                "category" => $fieldName,
                "value" => $fieldValue,
                "color" => getColorByFieldName($fieldName),
            ];
        }
    }
}
echo json_encode($json);

function getColorByFieldName($fieldName) 
{
    if ($fieldName === 'ONTIME' || $fieldName === 'cONTIME') {
        return '#0f9b0f';
    }
    if ($fieldName === 'LATE' || $fieldName === 'cLATE') {
        return '#dd1818';
    }
    if ($fieldName === 'TOTAL' || $fieldName === 'cTOTAL') {
        return '#0575E6';
    }
}

使用PHP从SQL查询创建JSON

1 个答案: