Does not return rows from serialized data by using like operator
Select * from `transactions` where 'order_id' like "'%115%'"
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Jignesh Joisar
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Rahul Jat
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9Show us your sample data and expected output. – Madhur Bhaiya Oct 24 '18 at 06:19
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in your table order_id is a string or an integer? – Sfili_81 Oct 24 '18 at 06:40
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Please check this thread: https://stackoverflow.com/questions/24508164/how-do-i-search-from-serialize-field-in-mysql-database – Sami Ahmed Siddiqui Oct 24 '18 at 07:33
2 Answers
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Looking to your code you are trying to match not the coumn name but the string 'order_id' with the string "'%115%'" that, obviuosly , don't match so you have no rows result ..then
avoid single quote around column name and use concat for form a proper like condition
Select * from `transactions` where order_id like concat('%', '115', '%')
and be sure your order_id is string data type columns otherwise you should use
Select * from `transactions` where order_id =115
ScaisEdge
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Use one of the following :
1) If your order_id field is integer and you want to get exact value try this
SELECT * FROM `transactions` WHERE order_id = 115
2) If your order_id field is integer and you want to get matching value try this
SELECT * FROM `transactions` WHERE CAST(order_id AS CHAR) LIKE '%115%'
3) If your order_id field is string and stored as serialized data try this
SELECT * FROM table WHERE field REGEXP '.*"array_key";s:[0-9]+:".array_value.".*'
ex. : SELECT * FROM `transactions` WHERE order_id REGEXP '.*"m5";s:[0-9]+:"115".*'
4) If your order_id field stored as simple string try this
SELECT * FROM `transactions` WHERE order_id LIKE '%115%'
I hope this will help you :)