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執行下面的PHP代碼中的錯誤appreas PHP代碼:
while($row = mysqli_fetch_assoc($query)){Whatever code}
:
<?php
require "init.php";
$Date = [];
$Subject = [];
$Desc= [];
$query = mysqli_query($con,"SELECT date, Subject, Desc FROM sherif_DCOAn");
while($row = mysqli_fetch_assoc($query)){
$Date[] = $row['date'];
$Subject[] = $row['Subject'];
$Desc[] = $row['Desc'];
echo json_encode($Date).','.json_encode($Subject).','.json_encode($Desc).',';
}
?>
當我添加的,而部分出現錯誤
我在一個不同的選擇上應用了它,它的工作原理如下:
$query = mysqli_query($con,"SELECT DISTINCT SiteName FROM CAB");
第一個錯在哪?
你救了我:) ..這是答案 –
@SheriffSaidElahl我去過那裏的次數超過了我的數量!稍微闡述了答案,謝謝你的最佳答案。 –
這些是列名,而不是表。 – jkavalik