Business
Jobs
  • About Us
  • Solutions
    • Job Postings
      Post your job and receive qualified candidates in 48h.
    • Candidate Assessments
      500+ technical and psychological tests, plus anti-fraud.
    • Headhunting
      Tailor-made executive search from start to finish.
    • Payroll + EOR
      Payroll dispersal and EOR across 15+ LATAM countries.
  • Pricing
  • Jobs

0

247
Views
Eliminar mediante programación una matriz de objetos, JavaScript

Estoy tratando de recordar mediante programación una matriz de objetos, pero me faltan habilidades y tampoco parece que encuentre respuestas en la web.

Por lo tanto, quiero poder, mientras dibujo un cardObj, eliminar su alternativa inversa (al revés). Tenga en cuenta que quiero hacer lo mismo si la versión inversa es el cajón primero: retire la vertical de la plataforma.

Aquí está el código:

¡Ejercicio Cardz!

¡EJERCICIO!

Pasado presente Futuro
Untado

PASADOPRESENTEFUTURO// -----------JAVASCRIPT--------------//
 // CARDS OBJECT ------------------------------------------------------------------------------------> let cardObj = [ {name: "0_Fool", imgUrl: "data/0_Fool.jpg"}, {name: "0_Fool_R", imgUrl: "data/0_Fool_R.jpg"}, //{name: "1_Magician", imgUrl: "data/1_Magician.jpg"}, //{name: "1_Magician_R", imgUrl: "data/1_Magician_R.jpg"}, ]; //--- ------------------------------------------------------------------------------------> //SHUFFLE CARDS ------------------------------------------------------------------------------------> function shuffle(array){ let currentIndex = array.length, randomIndex; while(currentIndex != 0){ randomIndex = Math.floor(Math.random() * currentIndex); currentIndex--; [array[currentIndex], array[randomIndex]] = [array[randomIndex], array[currentIndex]]; } return array; }; shuffle(cardObj); //--- ------------------------------------------------------------------------------------> //DRAW ------------------------------------------------------------------------------------> function drawCard(id){ if(cardObj.length === 0){ return; } let castCard = document.getElementById(id); let card = document.createElement('img'); card.setAttribute('src', cardObj[0].imgUrl); card.setAttribute('height', '272'); card.setAttribute('width', '185'); card.innerHTML = cardObj[0].imgUrl; if(castCard.id === 'imgPast'){ document.getElementById("btnPast").replaceChild(card, castCard); }else if(castCard.id === 'imgPresent'){ document.getElementById("btnPresent").replaceChild(card, castCard); }else if(castCard.id === 'imgFuture'){ document.getElementById("btnFuture").replaceChild(card, castCard); } if(cardObj[0].name === cardObj[0].name){ cardObj = cardObj.filter(function(f) {return f !== cardObj.name + "_R"}); } cardObj.shift(); return false; } //--- ----------------------------------------------------------------------------------</script></body></html>
about 4 years ago · Juan Pablo Isaza
1 answers
Answer question

0

Estoy bastante confundido acerca de su pregunta aquí, pero asumiré que lo que quiere es eliminar el elemento "opuesto" de la matriz según la carta que se extraiga.

 let cardObj = [ {name: "0_Fool", imgUrl: "data/0_Fool.jpg"}, {name: "0_Fool_R", imgUrl: "data/0_Fool_R.jpg"}, {name: "1_Magician", imgUrl: "data/1_Magician.jpg"}, {name: "1_Magician_R", imgUrl: "data/1_Magician_R.jpg"}, ]; function drawCard(id){ if(cardObj.length === 0) return; // assuming that all 'reverse' cards end with '_R' and that 'id' is the 'name' const isReverse = id.endsWith('_R'); // construct opposite card ID const oppositeCardID = isReverse ? id.substring(0, id.lastIndexOf('_')) : (id + '_R'); // Find opposite card index in the array const oppositeCardIndex = cardObj.findIndex(card => card.name == oppositeCardID); if (oppositeCardIndex < 0) return; // opposite card not found // Remove opposite card from deck cardObj.splice(oppositeCardIndex, 1); } console.log("Cards before drawing:", JSON.stringify(cardObj)); console.log("Drawing card 1_Magician"); drawCard("1_Magician"); console.log("Cards after drawing:", JSON.stringify(cardObj));

about 4 years ago · Juan Pablo Isaza Report
Answer question
Find remote jobs

Discover the new way to find a job!

Top jobs
Top job categories
Business
Post vacancy Pricing Sales
Legal
Terms and conditions Privacy policy
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Show me some job opportunities
There's an error!