问题描述
我正在使用以下内容将数据库中的数据返回到我的视图.数据库列包含以下数据:
I'm using the following to get data from a database back to my view. The database column has the following data:
'hello'
'howdy'
'blablabl'
但是,当我使用下面的php并在jQuery ajax成功调用中对其进行检索时,它会以字符串形式获取文字数组,因此当我执行以下操作时:
however, when I use the php below, and retrieve it inside jQuery ajax success call, its getting a literal array as a string, so when I do the following:
success: function(data){
for (var i = 0; i < data.length; i++) {
var comment = data[i];
data[i]
是'[', '"', 'h', 'e','l','l'
等...如何避免这种情况
data[i]
is '[', '"', 'h', 'e','l','l'
etc... how can I avoid this
if ($result = $mysqli->query("SELECT * FROM comments")) {
$row_cnt = $result->num_rows;
if ($row_cnt > 0) {
while($row = $result->fetch_assoc()) {
$array[] = $row["comment"];
}
echo json_encode($array);
} else {
echo "no data";
}
$result->close();
}
推荐答案
由于要从php传递JSON字符串,因此需要首先将其解析为javascript中的对象.
Since you are passing a JSON string from the php, you need to first parse it into an object in your javascript.
您可以按如下所示进行操作-
You can do this as shown -
obj = JSON.parse(data);
然后您可以从obj获取您的值.
You can then get your value from the obj.
这篇关于服务器回显字母,而不是整个单词的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!