PHP mysqli的语句错误

而与PHP mysqli的语句错误

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执行下面的PHP代码中的错误appreas PHP代码：

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).','; 
} 
?>

当我添加的，而部分出现错误

我在一个不同的选择上应用了它，它的工作原理如下：

$query = mysqli_query($con,"SELECT DISTINCT SiteName FROM CAB");

第一个错在哪？

来源

2016-05-15 Sheriff Said Elahl

您的查询，SELECT date, Subject, Desc FROM sherif_DCOAn包含MySQL reserved keywords。列名为date,Subject和Desc。

你可以BACKTICK那些列名逃脱他们喜欢：

SELECT `date`, `Subject`, `Desc` FROM sherif_DCOAn

，你应该是不错的。仅供将来参考，我相当确定，建议不要在名称中使用这些关键字。

来源

2016-05-15 12:09:14

你救了我:) ..这是答案 –

@SheriffSaidElahl我去过那里的次数超过了我的数量！稍微阐述了答案，谢谢你的最佳答案。 –

这些是列名，而不是表。 – jkavalik

-1

改变这一点：

$query = mysqli_query($con,"SELECT date, Subject, Desc FROM sherif_DCOAn");

到：

$query = mysqli_query($con,"SELECT * FROM sherif_DCOAn");

你在这里呼唤你的结果

：

while($row = mysqli_fetch_assoc($query)){ 
$Date[] = $row['date']; 
$Subject[] = $row['Subject']; 
$Desc[] = $row['Desc'];

与其他词，你两次打电话给你行。如果你明白我的意思...

来源

2016-05-15 12:22:47

PHP mysqli的语句错误

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