在 php 我有這個數組$MAtoJS[] = array('Max','Mustermann','N','A80');$MAtoJS[] = array('Michaela','May','N','A78');$MAtoJS[] = array('Hans','Gerstelhuber','N','M12');$MAtoJS[] = array('Alfred E.','Neumann','N','T25');$MAtoJS[] = array('James','Bond','N','M72');仍然在 php 部分我為 js 準備了數組$MAarrayForJS = json_encode(json_encode($MAForJS));在 javascript 部分，我創建了一個 js-arrayvar MAarray = new Array(<?php echo $MAarrayForJS; ?>); alert(MAarray)MAarray 的內容是[["Max","Mustermann","N","A80"],["Michaela","May","N","A78"],["Hans","Gerstelhuber","N","M12"],["Alfred E.","Neumann","N","T25"],["James","Bond","N","M72"]]使用console.log(MAarray[0]);例如，我試圖通過此代碼獲取 Hans 的名字var FirstName = MAarray[0][2][1];這導致 console.log 中的“未定義”。如何訪問特定數組中的特定值，在這種情況下，將 var FirstName 設置為第三個“子數組”中的值“Hans”？
3 回答
溫溫醬
TA貢獻1752條經驗 獲得超4個贊
請注意，數組是2 維而不是3 維，您訪問的第 3 維將始終產生未定義，請查看下面的代碼片段，其中獲取值，并根據要求進行更改：
let MAarray = [["Max","Mustermann","N","A80"],["Michaela","May","N","A78"],["Hans","Gerstelhuber","N","M12"],["Alfred E.","Neumann","N","T25"],["James","Bond","N","M72"]];
console.log(MAarray[2][0]);
MAarray[2][0] = "New Name";
console.log(MAarray[2][0]);
慕虎7371278
TA貢獻1802條經驗 獲得超4個贊
你在找這個……？
$MAtoJS = [];
$MAtoJS[] = array('Max','Mustermann','N','A80');
$MAtoJS[] = array('Michaela','May','N','A78');
$MAtoJS[] = array('Hans','Gerstelhuber','N','M12');
$MAtoJS[] = array('Alfred E.','Neumann','N','T25');
$MAtoJS[] = array('James','Bond','N','M72');
$MAarrayForJS = json_encode($MAtoJS);
?>
<script>
var MAarray = <?php echo $MAarrayForJS; ?>;
console.log(MAarray[0][0]); // Max
</script>
- 3 回答
- 0 關注
- 225 瀏覽
添加回答
舉報
0/150
提交
取消