Flask到外部javascript变量传输失败？

Question

我想将app.py中的文件名解析为graph.js。但我无法获取js文件的变量名称。我正在使用烧瓶

这是文件结构

- app/
  - flask_app.py
  - templates/
    -> index.html
    -> data/
      -> file1.csv
  - static/
    -> js/
      -> create-graph.js

我的应用看起来像这样（相关部分）：

data_files = ["file1.xlsx", "file2.xlsx", "file3.xlsx"]
cur = 0
df1 = pd.read_excel("data/" + data_files[cur])

df1.to_csv('templates/data/' + data_files[cur][:10] + '.csv', encoding='utf-8', index=False)

所以我有从xlsx转换为csv的文件，然后我想用它的名字拉出csv文件。我需要将data_files和cur传递给我的.js文件... 以下是相关的.js：

function parseData(createGraph) {
var name = '{{data_files[cur][:10]}}';
console.log(name);
Papa.parse("static/data/" + name + ".csv", {
    download: true,
    complete: function(results) {
        createGraph(results.data);
    }
});

}

但似乎我无法将文件名传递到此处：var name = '{{data_files[cur][:10]}}';

在控制台上，它为我提供完全相同的字符串

那么我做错了什么？

我也尝试将其添加到app：

@app.route('/create-graph.js')
def script():
    return render_template('create-graph.js', fileName="data_files[cur][:10]")

仍然没有运气......

如果有人问，这里我如何调用.js文件：

...
<script src="{{ url_for('static', filename='js/create-graph.js') }}"></script>
...

Answer 1

You need to pass the parameter to render_template as a variable with desired values.

If you've only one file then you can just use.

app.py

@app.route('/create-graph.js')
def script():
    data_file = "test.csv"
    return render_template('create-graph.js', file_name=data_file)

.js file

function parseData(createGraph) {
var name = '{{file_name}}';
console.log(name);
Papa.parse("static/data/" + name + ".csv", {
    download: true,
    complete: function(results) {
        createGraph(results.data);
    }
});
}

If data_files is a list of list or dictionary of the list.

.js

function parseData(createGraph) {
var name = '{{data_files_cur[:10]}}'; // here you can get at max 10 files
console.log(name);
Papa.parse("static/data/" + name + ".csv", {
    download: true,
    complete: function(results) {
        createGraph(results.data);
    }
});
}

app.py

@app.route('/create-graph.js')
def script():
    # define cur  and data_files
    data_files_cur = data_files[cur];
    return render_template('create-graph.js', data_files_cur=data_files_cur)

It looks you don't know much about flask template engine syntax. I would recommend you should read about Jinja/Mako or Django template syntax.