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"16": [1994, 53, 61], "17": [1993, 75, 87], "length": 18, "callee": argumentsToArray };
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Inserting or Deleting Elements from an Array
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Now then, say the elements for the 2000 2002 seasons are invalid inasmuch as the wins and losses are reversed, and 2002 is in there twice. To delete those four invalid elements and insert new ones in their place, we can invoke splice() on pirates. The first parameter is the index of the first element to delete, so 8 for the 2002 season. The second parameter is the number of elements to delete, which would be 4. From there, any additional parameters are elements to splice into the array. Note that it s OK to add more or fewer elements than you deleted; JavaScript will keep the array indexes sequential behind the scenes.
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So if we add three elements in place of the four invalid ones, the indexes in pirates remain sequential, ordered 0 to 17 as Figure 6 39 displays. var pirates = [[2010, 57, 105], [2009, 62, 99], [2008, 67, 95], [2007, 68, 94], [2006, 67, 95], [2005, 67, 95], [2004, 72, 89], [2003, 75, 87], [2002, 89, 72], [2002, 89, 72], [2001, 100, 62], [2000, 93, 69], [1999, 78, 83], [1998, 69, 93], [1997, 79, 83], [1996, 73, 89], [1995, 58, 86], [1994, 53, 61], [1993, 75, 87]]; pirates.splice(8, 4, [2002, 72, 89], [2001, 62, 100], [2000, 69, 93]); console.dir(pirates);
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Figure 6 39. Deleting and inserting elements into an array with splice() Now, say the 2000 2002 seasons are simply missing. We can use splice() to insert those. That is to say, splice() does not require you to delete any elements. So if we pass 0 for the second parameter to splice(), JavaScript will insert the three new elements, and, as before, keep the indexes sequential behind the scenes, as Figure 6 40 displays. var pirates = [[2010, 57, 105], [2009, 62, 99], [2008, 67, 95], [2007, 68, 94], [2006, 67, 95], [2005, 67, 95], [2004, 72, 89], [2003, 75, 87], [1999, 78, 83], [1998, 69, 93], [1997, 79, 83], [1996, 73, 89], [1995, 58, 86], [1994, 53, 61], [1993, 75, 87]]; pirates.splice(8, 0, [2002, 72, 89], [2001, 62, 100], [2000, 69, 93]); console.dir(pirates);
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Figure 6 40. There s no need to delete elements prior to inserting new ones with splice(). Conversely, splice() can be used just to delete elements. That is to say, we re not required to insert new elements in place of the ones we delete. So say pirates contains elements for the 1990 92 seasons and we want to delete those so that only the seasons comprising the 18-year losing streak remain. To do so, we d simply pass 18 and 3 to splice(). JavaScript will then clip those three winning seasons from pirates, as Figure 6 41 displays. var pirates = [[2010, 57, 105], [2009, 62, 99], [2008, 67, 95], [2007, 68, 94], [2006, 67, 95], [2005, 67, 95], [2004, 72, 89], [2003, 75, 87], [2002, 72, 89], [2001, 62, 100], [2000, 69, 93], [1999, 78, 83], [1998, 69, 93], [1997, 79, 83], [1996, 73, 89], [1995, 58, 86], [1994, 53, 61], [1993, 75, 87], [1992, 96, 66],
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[1991, 98, 64], [1990, 95, 67]]; pirates.splice(18, 3); console.dir(pirates);
Figure 6 41. There s no need to replace deleted elements with new ones when invoking splice(). Even more succinctly, we could accomplish the same thing by simply passing 18. JavaScript will then delete any elements from that index all the way to the end of the array. So, as Figure 6 42 displays, passing 18 modifies pirates, just as passing 18 and 3 did. var pirates = [[2010, 57, 105], [2009, 62, 99], [2008, 67, 95], [2007, 68, 94], [2006, 67, 95], [2005, 67, 95], [2004, 72, 89], [2003, 75, 87], [2002, 72, 89], [2001, 62, 100], [2000, 69, 93], [1999, 78, 83], [1998, 69, 93], [1997, 79, 83], [1996, 73, 89], [1995, 58, 86], [1994, 53, 61],
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