“py sqlite si l'entrée existe” Réponses codées

Si sorse python sql

for name in ('bar','foo'): 
    cursor.execute("SELECT rowid FROM components WHERE name = ?", (name,))
    data=cursor.fetchall()
    if len(data)==0:
        print('There is no component named %s'%name)
    else:
        print('Component %s found with rowids %s'%(name,','.join(map(str, next(zip(*data))))))

Faithful Flamingo

py sqlite si l'entrée existe

-- Count entries in database
SELECT count(*);

-- Count rows in table
SELECT count(*) FROM peoples;
-- returns the number of rows in this table

-- Count rows with filter
SELECT count(*) FROM peoples WHERE age = 20;
-- returns the number of rows/peoples with an age of 20

-- If none (matching) row is in the specified
-- table it will return 0. So you can
-- check for "if (SQL_return >= 1)"

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