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Home/ Questions/Q 8360625
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T11:30:54+00:00 2026-06-09T11:30:54+00:00

When I define a model TeMdl and then use Fn TeMdl.load(1),it will send a

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When I define a model “TeMdl” and then use Fn TeMdl.load(1),it will send a request with param id=1.
so..How to Change the id param so the request can like this:”……\?uid=1″?
i am a fresh to extjs!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T11:30:55+00:00Added an answer on June 9, 2026 at 11:30 am

    From the comments in the Ext documentation:

    The load() method will not honor the idProperty of the Model, and will assume it’s “id”.

    The following subclass fixes this behaviour by using idProperty instead of id.

    Ext.define("Ux.data.Model", {
      extend: "Ext.data.Model",
      statics: {
        load: function(id, config){
          config = Ext.apply({}, config);
    
          var params={};
          params[this.prototype.idProperty] = id;
    
          config = Ext.applyIf(config, {
              action: 'read',
              params: params
          });
    
          var operation  = Ext.create('Ext.data.Operation', config),
              scope      = config.scope || this,
              record     = null,
              callback;
    
          callback = function(operation) {
              if (operation.wasSuccessful()) {
                  record = operation.getRecords()[0];
                  Ext.callback(config.success, scope, [record, operation]);
              } else {
                  Ext.callback(config.failure, scope, [record, operation]);
              }
              Ext.callback(config.callback, scope, [record, operation]);
          };
    
          this.proxy.read(operation, callback, this);
        }
      }
    });
    
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