Home > Web Front-end > JS Tutorial > body text

How do I parse server-side errors and display them in a dialog box using jqGrid?

Linda Hamilton
Release: 2024-11-14 18:50:03
Original
641 people have browsed it

How do I parse server-side errors and display them in a dialog box using jqGrid?

Handling Server-Side Errors and Validation in jqGrid

Question:

In my JSON responses, I have "STATUS" and "errors" properties. How can I utilize these properties in jqGRid to parse errors and display them in a dialog box?

Answer:

The key to handling errors with jqGrid is to follow HTTP protocol rules. Successful responses include a status code of 200, while error responses have specific status codes (e.g., 404 Not Found).

Error Handling Implementation:

In your loadError event handler, you can check the HTTP status code and parse the response accordingly:

loadError: function (jqXHR, textStatus, errorThrown) {
    if (jqXHR.status === 404) {
        // Error handling for 404 Not Found
    } else if (typeof jqXHR.responseText === "string") {
        // Error handling based on the JSON response
        var errorInfo = $.parseJSON(jqXHR.responseText);
        var errorMessages = "";
        for (var i = 0; i < errorInfo.length; i++) {
            errorMessages += errorInfo[i].Source + ": " + errorInfo[i].Message;
        }
        alert("Error:\n" + errorMessages);
    }
}
Copy after login

Note: You can customize the error message display to match your application's needs using HTML or CSS.

Additional Considerations:

  • If you send errors as JSON, ensure the Content-Type header is set to application/json.
  • Free jqGrid also provides a default loadError implementation that displays a readable error message in an error div above the grid.
  • You can use the displayErrorMessage method to set custom error messages in the error div.

The above is the detailed content of How do I parse server-side errors and display them in a dialog box using jqGrid?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template