解析错误(T_VARIABLE)

时间:2014-10-20 22:12:05

标签: php mysql mysqli explode parse-error

我在爆炸线上得到一个解析错误。 我只想把我已经存储在我的数据库中的关键字(brad pitt,fight club,project mayhem)转换成用逗号分解的阵列。然后将该数组与新用户输入合并。

<?php
$tags = $_POST['tagArray'];
$title = $_POST['title'];

$DBServer = "localhost";
$DBUser = "------";
$DBPass = "-------";
$DBName = "movies";

$conn = mysqli_connect($DBServer, $DBUser, $DBPass, $DBName);


$query = "SELECT keywords FROM movieTitles WHERE movie='" .$title. "'";
$result = mysqli_query($conn, $query);
$row = mysqli_fetch_array($result);
$keywords = $row['keywords'];

$current = explode("," $keywords[0]);
$array = array_unique(array_merge($tags,$current));

echo $array;

mysqli_close($conn);

?>

3 个答案:

答案 0 :(得分:1)

$current = explode(",", $keywords[0]);

你错过了一个额外的逗号

答案 1 :(得分:0)

您错过了分隔符和文本之间的逗号

$current = explode("," , $keywords[0]);

答案 2 :(得分:0)

$current = explode("," $keywords[0]);

您错过了逗号,以分隔函数参数

应该是:

$current = explode(",", $keywords[0]);

但除了您的错误,$row['keywords'];是一个字符串,因此使用[0];将搜索字符串的第一个字符。例如:

$row['keywords'] = "Test";
echo $row['keywords'][0]; // Output T

总的来说,这样做最好:

$current = explode(",", $keywords);