Mixing
Cant get Data out of then() Scrapeit | NODE JS
-2
This is my simple App
for(var i = 0 ; i < prod.length;i++ ){
var urls = prod[i].URL
urls = urls.split(" ")[0];
scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
urli = "https://www.choithrams.com"+data.avatar
var name = prod[i].Name
var price = prod[i].price
console.log(name)
})
}
I want to get the result step by step but it is not iterating correctly.
Secondly I, for example, I have a global variable outside and I want to overwrite its value from the one I get, I cannot simply do it.
the Result is
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
which is the last item in mineJSON.
I am a beginner in all this kindly do help.
node.js
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
add a comment |
-2
This is my simple App
for(var i = 0 ; i < prod.length;i++ ){
var urls = prod[i].URL
urls = urls.split(" ")[0];
scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
urli = "https://www.choithrams.com"+data.avatar
var name = prod[i].Name
var price = prod[i].price
console.log(name)
})
}
I want to get the result step by step but it is not iterating correctly.
Secondly I, for example, I have a global variable outside and I want to overwrite its value from the one I get, I cannot simply do it.
the Result is
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
which is the last item in mineJSON.
I am a beginner in all this kindly do help.
node.js
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Don't usevar. Useconst(or, when not possible,let)
– CertainPerformance
Jan 1 at 11:42
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55
add a comment |
-2
-2
-2
This is my simple App
for(var i = 0 ; i < prod.length;i++ ){
var urls = prod[i].URL
urls = urls.split(" ")[0];
scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
urli = "https://www.choithrams.com"+data.avatar
var name = prod[i].Name
var price = prod[i].price
console.log(name)
})
}
I want to get the result step by step but it is not iterating correctly.
Secondly I, for example, I have a global variable outside and I want to overwrite its value from the one I get, I cannot simply do it.
the Result is
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
which is the last item in mineJSON.
I am a beginner in all this kindly do help.
node.js
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
This is my simple App
for(var i = 0 ; i < prod.length;i++ ){
var urls = prod[i].URL
urls = urls.split(" ")[0];
scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
urli = "https://www.choithrams.com"+data.avatar
var name = prod[i].Name
var price = prod[i].price
console.log(name)
})
}
I want to get the result step by step but it is not iterating correctly.
Secondly I, for example, I have a global variable outside and I want to overwrite its value from the one I get, I cannot simply do it.
the Result is
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
Mara Chopped Tomatoes Easy Open 400 gms
which is the last item in mineJSON.
I am a beginner in all this kindly do help.
node.js
node.js
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
edited Jan 1 at 11:44
Raja Osama
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
asked Jan 1 at 11:40
Raja OsamaRaja Osama
36
36
Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Don't usevar. Useconst(or, when not possible,let)
– CertainPerformance
Jan 1 at 11:42
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55
add a comment |
Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Don't usevar. Useconst(or, when not possible,let)
– CertainPerformance
Jan 1 at 11:42
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55
Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Don't use
var. Use const (or, when not possible, let)– CertainPerformance
Jan 1 at 11:42
Don't use
var. Use const (or, when not possible, let)– CertainPerformance
Jan 1 at 11:42
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55
add a comment |
1 Answer
1
active
oldest
votes
0
It's better using Promise.all in this case:
Promise.all(prod.map(({ URL, Name, price }) => {
const URL = URL.split(" ")[0];
return scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
const urli = "https://www.choithrams.com" + data.avatar
return {
urli, Name, price
}
})
})).then(results => {
// Results is an array of:
// { urli: "...", Name: "...", price: ... }
...
})
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
It's better using Promise.all in this case:
Promise.all(prod.map(({ URL, Name, price }) => {
const URL = URL.split(" ")[0];
return scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
const urli = "https://www.choithrams.com" + data.avatar
return {
urli, Name, price
}
})
})).then(results => {
// Results is an array of:
// { urli: "...", Name: "...", price: ... }
...
})
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
add a comment |
0
It's better using Promise.all in this case:
Promise.all(prod.map(({ URL, Name, price }) => {
const URL = URL.split(" ")[0];
return scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
const urli = "https://www.choithrams.com" + data.avatar
return {
urli, Name, price
}
})
})).then(results => {
// Results is an array of:
// { urli: "...", Name: "...", price: ... }
...
})
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
add a comment |
0
0
0
It's better using Promise.all in this case:
Promise.all(prod.map(({ URL, Name, price }) => {
const URL = URL.split(" ")[0];
return scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
const urli = "https://www.choithrams.com" + data.avatar
return {
urli, Name, price
}
})
})).then(results => {
// Results is an array of:
// { urli: "...", Name: "...", price: ... }
...
})
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
It's better using Promise.all in this case:
Promise.all(prod.map(({ URL, Name, price }) => {
const URL = URL.split(" ")[0];
return scrapeIt(urls, {
avatar: {
selector: ".image img"
, attr: "src"
}
}).then(async ({ data, response }) => {
// console.log(`Status Code: ${response.statusCode}`)
const urli = "https://www.choithrams.com" + data.avatar
return {
urli, Name, price
}
})
})).then(results => {
// Results is an array of:
// { urli: "...", Name: "...", price: ... }
...
})
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
answered Jan 1 at 17:40
Ionică BizăuIonică Bizău
58.7k57200348
58.7k57200348
add a comment |
add a comment |
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Possible duplicate of JavaScript closure inside loops – simple practical example
– CertainPerformance
Jan 1 at 11:42
Don't use
var. Useconst(or, when not possible,let)– CertainPerformance
Jan 1 at 11:42
@CertainPerformance I have gotten this sequence of iteration ibb.co/Qf5fKxC
– Raja Osama
Jan 1 at 11:55