Question

如何通过php创建var javascript？

<?PHP
include("connect.php");
$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
while($resilt_row = mysqli_fetch_array($get_data))
{ 
    $bad_words = $bad_words."".$resilt_row [word].",";
}
//echo $bad_words;
?>

<script>
var bad_words = ["<?PHP echo $bad_words; ?>"];
alert(bad_words);
</script>

我想得到像这样的{j}这个var bad_words = ["fuck", "ass"]; 警报时，它只会得到空白结果。

我该怎么做？

Answer 1

<?php

include("connect.php");
$query = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
$badWords = [];

while($row = mysqli_fetch_array($query))
{ 
    $badWords[] = $row['word'];
}

JS

<script>
var bad_words = <?= json_encode($badWords); ?>;
alert(bad_words[0]);
console.log(bad_words);
</script>

如果它是一个数组，我不喜欢粘合字符串然后将其作为具有json_encode()函数的数组传递。同样在视图中，使用<?= ?>标记

的短语法会更好

Answer 2

<?PHP
include("connect.php");
$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
while($resilt_row = mysqli_fetch_array($get_data))
{ 
 $bad_words[] = $resilt_row['word'];
}
//echo $bad_words;
?>

<script>
var bad_words = ["<?php print implode('","',$bad_words); ?>"];
alert(bad_words);
</script>

Answer 3

如果你想把它们警告为一个字符串，你可以这样做：

<script>
    var bad_words = <?php echo json_encode($bad_words) ?>;
    console.log(bad_words);
</script>

否则，您可以将它们打印为数组：

Easy Install. $ sudo apt-get install python-setuptools python-dev build-essential  
pip. $ sudo easy_install pip
virtualenv. $ sudo pip install --upgrade virtualenv.
python3-dev tools.$sudo apt-get install python3-dev

Answer 4

您正在制作的琴弦不是您想要的琴弦你没有添加引号。

这样做，使用数组更好：

$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
$bad_words = array();
while($resilt_row = mysqli_fetch_array($get_data))
{ 
    $bad_words[] = "'" . $resilt_row[word]. "'";
}

然后，在你的js中：

var bad_words = ["<?php echo implode(",", $bad_words) ?>"];

或（简称）

var bad_words = ["<?= implode(",", $bad_words) ?>"];

如何通过PHP创建var javascript？

4 个答案: