Javascript console.log()返回对php字符串的引用错误
我试图将数组值输出到控制台中,但出现未定义的错误: 错误Javascript console.log()返回对php字符串的引用错误,javascript,php,mysql,Javascript,Php,Mysql,我试图将数组值输出到控制台中,但出现未定义的错误: 错误 ReferenceError: Ciencia is not defined 代码 $sql = "SELECT topic FROM book_list WHERE topic != '' ORDER BY topic ASC"; $result = mysqli_query($db, $sql); <?php while($topic = mysqli_fetch_a
ReferenceError: Ciencia is not defined
代码
$sql = "SELECT topic FROM book_list WHERE topic != '' ORDER BY topic ASC";
$result = mysqli_query($db, $sql);
<?php
while($topic = mysqli_fetch_assoc($result)){?>
console.log(<?php echo $topic['topic']; ?>);
<?php }?>
谢谢 似乎它在尝试
console.log(Ciencia)
而不是console.log(“Ciencia”)
将您的线路更改为:
console.log("<?php echo $topic['topic']; ?>");
console.log(“”);
似乎它正在尝试console.log(Ciencia)
而不是console.log(“Ciencia”)
将您的线路更改为:
console.log("<?php echo $topic['topic']; ?>");
console.log(“”);
您需要将PHP字符串转换为Javascript字符串,而不是尝试将PHP字符串本身用作表达式json_encode()
非常适合这一点
console.log(<?php echo json_encode($topic['topic']); ?>);
console.log();
如果PHP变量包含需要编码的特殊字符(例如引号),这比在echo语句周围加引号要好。您需要将PHP字符串转换为Javascript字符串,而不是尝试将PHP字符串本身用作表达式
json_encode()
非常适合这一点
console.log(<?php echo json_encode($topic['topic']); ?>);
console.log();
如果PHP变量包含需要编码的特殊字符(例如引号),这比在echo语句周围加引号要好。阅读生成的源代码,您就会看到问题阅读生成的源代码,您就会看到问题