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3.11 Sorting an Array of ObjectsNN 4, IE 4 3.11.1 ProblemYou want to sort an array of objects based on the value of one of the properties of the objects. 3.11.2 SolutionSorting an array of objects relies on a logical extension of the comparison function described for simple arrays in Recipe 3.5. Define a comparison function as usual, but let the actual comparisons work on the properties of the objects being passed to the function two at a time. To demonstrate the concept, we'll start with the array of sales objects: var sales = new Array( ); sales[sales.length] = {period:"q1", region:"east", total:2300}; sales[sales.length] = {period:"q2", region:"east", total:3105}; ... sales[sales.length] = {period:"q4", region"west", total:3810}; If you want to sort the sales array in descending order of the values of the total properties of each object, define a comparison function that returns the appropriate values based on the arithmetic: function compareTotals(a, b) { return b.total - a.total; } Because each array entry passed as parameters a and b is an object, you can use those parameter variables to reference the properties of the objects as they pass through in waves during the full sort operation. To sort the array by way of the comparison function, pass the function's reference to the sort( ) method of the array: sales.sort(compareTotals); Recall that the sort( ) method modifies the order of the original array. But you can invoke other sort( ) methods (that call other comparison functions) to resort the array by other criteria. 3.11.3 DiscussionComparison functions can get rather elaborate if necessary. It all depends on the kind of data in your object properties and what kind of sorting you need to perform. For example, if an object is defined with separate properties for month, day, and year, and if you want to sort the objects by the dates that those numbers represent, the comparison function can create date objects from those values and then compare the resulting date objects: function compareDates(a, b) { var dateA = new Date(a.year, a.month, a.date); var dateB = new Date(b.year, b.month, b.date); return dateA - dateB; } If sorting is required of string values in properties, you have to be more explicit in the comparisons you perform and the values you return. You may also want to eliminate case as a factor by comparing values converted to all upper- or all lowercase characters. The following function sorts string values of the lastName property of an array of objects: function compareNames(a, b) { var nameA = a.lastName.toLowerCase( ); var nameB = b.lastName.toLowerCase( ); if (nameA < nameB) {return -1} if (nameA > nameB) {return 1} return 0; } The return values from the function fall into the three categories described for all array sorting in Recipe 3.5. And because the toLowerCase( ) method of a string doesn't disturb the case of the original string, the object values are ready to be displayed as entered into the data structure. 3.11.4 See AlsoRecipe 3.5 for basic array sorting concepts; Recipe 14.16 for using object sorting to sort data for rendering in a Dynamic HTML table. |
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