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

175
Views
¿Por qué devolver el error 404? javascript para actualizar div

Estoy tratando de insertar un script para actualizar div con javascript/ajax. Encontré el siguiente código. No pongo la página prova.asp porque contiene solo una respuesta. Escriba "Hola" (luego modificaré con la consulta de la base de datos). Este código devuelve un error 404 y no entiendo por qué. La única URL es prova.asp y este archivo está en la misma carpeta.

¿Alguien puede decirme dónde me equivoco? el error es mio pero no se cual es

 <html> <head> <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script> <script type="text/javascript"> //If our user enters data in the username input, then we need to enable our button function OnChangedUsername(){ if(document.form1.newuserid.value == ""){ document.form1.btnCheckAvailability.disabled = true; } else { document.form1.btnCheckAvailability.disabled = false; } } function OnCheckAvailability(){ if(window.XMLHttpRequest){ oRequest = new XMLHttpRequest(); } else if(window.ActiveXObject) { oRequest = new ActiveXObject("Microsoft.XMLHTTP"); } oRequest.open("POST", "prova.asp", true); oRequest.onreadystatechange = UpdateCheckAvailability; oRequest.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); oRequest.send("strCmd=availability&strUsername=" + document.form1.newuserid.value); } function UpdateCheckAvailability(){ if(oRequest.readyState == 4){ if(oRequest.status == 200){ document.getElementById("Available").innerHTML = oRequest.responseText; } else { document.getElementById("Available").innerHTML = "Asynchronous Error"; } } } </script> </head> <body> <form method="post" action="javascript:void(0);" name="form1"> <table cellspacing="0"> <tr> <th><label for="newuserid">Username:</label></th> <td><input type="newuserid" name="newuserid" id="newuserid" size="20" onKeyUp="OnChangedUsername();"/></td> <td><input id="btnCheckAvailability" type="button" disabled="disabled" value="Check Availability" onClick="OnCheckAvailability();"></td> <td><div id="Available"></div></td> </tr> </table> </form> </body> </html>

about 4 years ago · Juan Pablo Isaza
1 answers
Answer question

0

Tal vez, el navegador intente ejecutar primero el script. Agregue su script antes de cerrar la etiqueta div .

 <body> ... ... <script .... > ..... </script> </body>

O intente agregar defer

 <body> ... ... <script .... defer> ..... </script> </body>

o async

 <body> ... ... <script .... async > ..... </script> </body>
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!