Insurance agencies in New Liskeard, Canada

List of Insurance agencies in New Liskeard, Ontario: find addresses, phone numbers, email, social networks, photos, customer reviews and more.

Popular insurance agencies in New Liskeard

0SELECT * FROM `locality` WHERE url = 'new-liskeard-on' LIMIT 1
1UPDATE locality SET weather = '{\"date\":\"2024-04-24\",\"data\":{\"2024-04-24\":{\"01\":{\"temp\":-5,\"pressure\":1016,\"humidity\":87,\"icon\":\"13n\",\"wind\":7},\"04\":{\"temp\":-9,\"pressure\":1020,\"humidity\":81,\"icon\":\"04n\",\"wind\":6},\"07\":{\"temp\":-11,\"pressure\":1026,\"humidity\":65,\"icon\":\"03d\",\"wind\":6},\"10\":{\"temp\":-5,\"pressure\":1027,\"humidity\":48,\"icon\":\"03d\",\"wind\":6},\"13\":{\"temp\":-1,\"pressure\":1026,\"humidity\":39,\"icon\":\"03d\",\"wind\":5},\"16\":{\"temp\":1,\"pressure\":1027,\"humidity\":41,\"icon\":\"02d\",\"wind\":4},\"19\":{\"temp\":-2,\"pressure\":1028,\"humidity\":57,\"icon\":\"03d\",\"wind\":2},\"22\":{\"temp\":-4,\"pressure\":1029,\"humidity\":70,\"icon\":\"03n\",\"wind\":1}},\"2024-04-25\":{\"01\":{\"temp\":-4,\"pressure\":1029,\"humidity\":70,\"icon\":\"03n\",\"wind\":1},\"04\":{\"temp\":-5,\"pressure\":1029,\"humidity\":70,\"icon\":\"01n\",\"wind\":2},\"07\":{\"temp\":-1,\"pressure\":1030,\"humidity\":55,\"icon\":\"01d\",\"wind\":2},\"10\":{\"temp\":5,\"pressure\":1029,\"humidity\":33,\"icon\":\"01d\",\"wind\":2},\"13\":{\"temp\":9,\"pressure\":1026,\"humidity\":26,\"icon\":\"02d\",\"wind\":3},\"16\":{\"temp\":11,\"pressure\":1025,\"humidity\":26,\"icon\":\"02d\",\"wind\":3},\"19\":{\"temp\":4,\"pressure\":1026,\"humidity\":57,\"icon\":\"01d\",\"wind\":1},\"22\":{\"temp\":0,\"pressure\":1027,\"humidity\":73,\"icon\":\"02n\",\"wind\":2}},\"2024-04-26\":{\"01\":{\"temp\":-1,\"pressure\":1028,\"humidity\":82,\"icon\":\"02n\",\"wind\":2},\"04\":{\"temp\":-1,\"pressure\":1028,\"humidity\":83,\"icon\":\"02n\",\"wind\":2},\"07\":{\"temp\":3,\"pressure\":1029,\"humidity\":64,\"icon\":\"02d\",\"wind\":2},\"10\":{\"temp\":10,\"pressure\":1028,\"humidity\":44,\"icon\":\"01d\",\"wind\":2},\"13\":{\"temp\":13,\"pressure\":1027,\"humidity\":36,\"icon\":\"01d\",\"wind\":4},\"16\":{\"temp\":14,\"pressure\":1025,\"humidity\":35,\"icon\":\"03d\",\"wind\":4},\"19\":{\"temp\":8,\"pressure\":1024,\"humidity\":60,\"icon\":\"04d\",\"wind\":2},\"22\":{\"temp\":5,\"pressure\":1024,\"humidity\":66,\"icon\":\"04n\",\"wind\":3}}}}' WHERE id = 16987 LIMIT 1
1SELECT * FROM `region` WHERE id = 176 LIMIT 1
0SELECT * FROM `category` WHERE url = 'insurance-agency' LIMIT 1
2SELECT locality_id, category_id, price_level FROM `category_item` WHERE locality_id = 16987 AND category_id = 132 GROUP BY locality_id, category_id, price_level
0SELECT COUNT(*) FROM `category_item` WHERE locality_id = 16987 AND category_id = 132
1SELECT * FROM category_item INNER JOIN item ON category_item.item_id = item.id WHERE category_item.locality_id = 16987 AND category_item.category_id = 132 ORDER BY item.name LIMIT 0,24
0SELECT * FROM `region` WHERE id IN ('176') LIMIT 1
0SELECT * FROM `locality` WHERE id IN ('16987') LIMIT 1
0SELECT * FROM `photos` WHERE `item_id` IN ('361633')
0SELECT item_id, category_id FROM `category_item` WHERE item_id IN ('361633')
0SELECT * FROM `category` WHERE id IN ('96','132') LIMIT 2
1SELECT locality_id, category_id FROM `category_item` WHERE locality_id = 16987 GROUP BY locality_id, category_id
0SELECT * FROM `category` WHERE `id` IN ('96','97','99','100','103','104','118','120','121','125','126','127','128','132','134','141','159','164') ORDER BY name
0SELECT * FROM counts where updated="2024-04-24" and table_name like 'news'
0SELECT * FROM counts where updated="2024-04-24" and table_name like 'news'
0SELECT * FROM `category` ORDER BY name
3SELECT * FROM `item` f JOIN ( SELECT RAND() * (SELECT MAX(id) FROM `item`) AS max_id ) AS m WHERE f.id >= m.max_id ORDER BY f.id ASC LIMIT 4
0SELECT * FROM `region` WHERE id IN ('176','174') LIMIT 2
0SELECT * FROM `locality` WHERE id IN ('16533','16812','16748','16353') LIMIT 4
0SELECT * FROM `photos` WHERE `item_id` IN ('32103','32104','32105','32106')
0SELECT * FROM `locality` f JOIN ( SELECT RAND() * (SELECT MAX(id) FROM `locality`) AS max_id ) AS m WHERE f.id >= m.max_id ORDER BY f.id ASC LIMIT 7
0SELECT * FROM `region` WHERE id IN ('168') LIMIT 1