NodeJs/MySQL Use a variable in where clause condition

Question

i want to use a variable in my Where clause like this :

con.query("SELECT * FROM vehicules WHERE "myvariable"=?",myothervariable,function (err, result)
    {
// some code
    });

it don't seems to work . is that even possible ?

EDIT

When i try this

con.query("SELECT * FROM vehicules WHERE 'myvariable'=?",myothervariable,function (err, result)

i got this as output

SELECT * FROM vehicules WHERE 'Modele'='X5'

Modele is my right column name but how can i remove the ' ' ?

what lib do you use with mysql?

Iurii Drozdov
– Iurii Drozdov

2017-10-16 09:16:25 +00:00
Commented Oct 16, 2017 at 9:16 — Iurii Drozdov
– Iurii Drozdov, Commented Oct 16, 2017 at 9:16
@IuriiDrozdov "mysql" is the lib's name

user8639911
– user8639911

2017-10-16 09:18:42 +00:00
Commented Oct 16, 2017 at 9:18 — user8639911
– user8639911, Commented Oct 16, 2017 at 9:18

Milan Velebit · Accepted Answer · 2017-10-16 10:17:08Z

1

Yep, it's possible indeed.

db.query("SELECT * FROM vehicles WHERE ??=?",[your_column_name,your_var_from_your_code],function(error,results){.......})

And if you want to use multiple statements, go for :

db.query("SELECT * FROM vehicles WHERE ??=? AND ??=?",[column_name_1,code_var_1,column_name_2,code_var2],function(error,results){...})

This should theoretically work, I haven't the time to actually test it.

edited Oct 16, 2017 at 10:17

answered Oct 16, 2017 at 9:18

Milan Velebit

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9 Comments

user8639911 Over a year ago

Not working , when i do like this it show me this error : field your_var unknow

Milan Velebit Over a year ago

Replace it with your real table column name, I placed random placeholders.. both your_column_name and your_var_from_your_code are placeholders, change them both.

user8639911 Over a year ago

But my column name is a variable , i do this query dynamically whit a dropdown list value .

Milan Velebit Over a year ago

I've edited my answer, you should clarify your question just a bit, we misunderstood you. Use double question marks (??) for identifiers and one(?) for values.

user8639911 Over a year ago

Thanks Working ! how could i make it work if i replace WHERE ??=? with WHERE like ??=? with contains (??=%?% don't work)

|

Iurii Drozdov · Accepted Answer · 2017-10-16 09:18:58Z

0

You could try this:

con.query("SELECT * FROM vehicules WHERE ?",{myvariable: myothervariable},function (err, result)
    {
// some code
    });

answered Oct 16, 2017 at 9:18

Iurii Drozdov

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1 Comment

user8639911 Over a year ago

Not working , field myvariable unknow (it's not taking my variable value but variable name)

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