variable in the sql nodejs

Question

Hello y need send variable in my request sql. For searching value in my database

var cent = "search";
con.connect(function (err) {
if (err) throw err;
var sql ="SELECT * FROM cadito.activitys WHERE description like ?";
//Send an array with value(s) to replace the escaped values:
con.query(sql, [%cen%], function (err, result) {
    console.log(global.re = JSON.stringify((result)));
});

});

My problème syntaxe error.

Thank for help me.

My log db_conn.js:20 con.query(sql, [%cent%], function (err, result) { ^

SyntaxError: Unexpected token '%' ?[90m at wrapSafe (internal/modules/cjs/loader.js:979:16)?[39m ?[90m at Module._compile (internal/modules/cjs/loader.js:1027:27)?[39m ?[90m at Object.Module._extensions..js (internal/modules/cjs/loader.js:1092:10)?[39m ?[90m at Module.load (internal/modules/cjs/loader.js:928:32)?[39m ?[90m at Function.Module._load (internal/modules/cjs/loader.js:769:14)?[39m ?[90m at Function.executeUserEntryPoint [as runMain] (internal/modules/run_main.js:72:12)?[39m ?[90m at internal/main/run_main_module.js:17:47?[39m

Boussadjra Brahim · Accepted Answer · 2020-12-03 10:07:34Z

1

Try out to use % inside the sql string and remove it from the params :

var cent = "search";
con.connect(function (err) {
if (err) throw err;
var sql ="SELECT * FROM cadito.activitys WHERE description like %?%";
//Send an array with value(s) to replace the escaped values:
con.query(sql, [cent], function (err, result) {
    console.log(global.re = JSON.stringify((result)));
});

});

answered Dec 3, 2020 at 10:07

Boussadjra Brahim

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Comments

Lamri Djamal · Accepted Answer · 2020-12-03 10:40:33Z

0

i found another issue in my question just placed % in my request and concaten my variable with %

var cent = "search";
con.connect(function (err) {
if (err) throw err;
var sql ="SELECT * FROM cadito.activitys WHERE description like ?";
//Send an array with value(s) to replace the escaped values:
con.query(sql, ['%'+cent+'%'], function (err, result) {
    console.log(global.re = JSON.stringify((result)));
});

});

working for me

answered Dec 3, 2020 at 10:40

Lamri Djamal

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