我编写了一个脚本,可以根据不同选项卡上的数据自动生成图表(单击新工作表上的按钮之后)。
我现在想添加一个“重置”按钮,该按钮将自动删除或删除生成的图表(以便我可以通过另一个按钮生成不同的图表)。
任何人都可以帮助我吗?我已经尝试了以下方法,但无济于事。...
import sqlite3
with sqlite3.connect(":memory:") as conn: # Using a context manager
c = conn.cursor()
c.execute("""
CREATE TABLE IF NOT EXISTS testing(
some_code INTEGER,
data TEXT)
""")
c.executemany("""
INSERT INTO testing VALUES (?, ?)
""", [[1, 'hi'], [2, 'bye'], [1, 'something']])
# Query the new database using a parameterized query
c.execute("select * from testing where some_code = ?", (1,))
if len(list(c)) > 0: # Exhausts the iterator and then throws the result away
print("Printing result set 1")
for row in c:
print(row)
print("End of result set 1")
print()
# Repeat the query
c.execute("select * from testing where some_code = ?", (1,))
print("Printing result set 2")
for row in c: # iterate the cursor
print(row)
print("End of result set 2")
print()
# And one more time but using fetchall()
c.execute("select * from testing where some_code = ?", (1,))
data = c.fetchall() # Exhaust the iterator but assign a list to a name
print("Printing result set 3")
for row in data:
print(row)
print("End of result set 3")
print()
# And we can keep on printing without re-querying
print("Printing result set 4")
for row in data:
print(row)
print("End of result set 4")
print()
答案 0 :(得分:0)
function delAllChartsFromSheet(name) {
var ss=SpreadsheetApp.getActive();
var sh=ss.getSheetByName(name);
var chts=sh.getCharts();
for(var i=0;i<chts.length;i++){
sh.removeChart(chts[i]);
}
}
答案 1 :(得分:0)
代码对我有用,我将其重写为:
function remove_charts_from_sheet(sheet_name) {
const tab = SpreadsheetApp.getActive().getSheetByName(sheet_name);
const charts = tab.getCharts();
for(let i = 0; i < charts.length; i++){
tab.removeChart(charts[i]);
}
}