Javascript File References Problem (with jQuery) -

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इससे पहले, मुझे यह था: 

  & lt; head & gt; & Lt; script src = "/ स्क्रिप्ट / jquery-1.3.2.min.js" प्रकार = "टेक्स्ट / जावास्क्रिप्ट" & gt; & lt; / स्क्रिप्ट & gt; & Lt; स्क्रिप्ट प्रकार = "टेक्स्ट / जावास्क्रिप्ट" & gt; Var optPrompt = "- एक का चयन करें -"; Var उपकैटेंज; Var पेरेंट कैट; Var next बीटीएन; Var मातृभावित = कार्य () {साफ़लेख (); यदि (this.selectedIndex == 0) {$ (subCats) .html ($ ("& lt; विकल्प & gt;")। पाठ (optPrompt)); } $ .getJSON ('& lt;% = Url.Action ("GetChildCategories")% & gt;', {parentId: $ (यह) .val ()}, फ़ंक्शन (डेटा) {subCats.options.length = 0; $ ("और" विकल्प " .text (this.Name) .appendTo (subCats);});}); } Var DisplayDescription = function (catId) {$ .ajax ({url: '& lt;% = Url.Action ("GetDescription")% & gt;', डेटा: {classId: catId}, डेटा प्रकार: 'html', सफलता: फ़ंक्शन (डेटा) {$ ("p # श्रेणी डिस्क्रिप्शन")। Html (डेटा);}}); } Var ChildChanged = function () {var cat से चयनित = this.selectedIndex! = 0; अगर (! बिल्ली सेलेक्टेड) ​​साफ़डिस्क्रिप्शन (); अन्य डिस्प्ले विवरण ($ (यह) .val ()); } Var ClearDescription = function () {$ ("p # श्रेणी डिस्क्रिप्शन")। Html (''); } $ (फ़ंक्शन () {parentCats = $ ("select # category")। Get (0); subcats = $ ("# subcategory चुनें")। Get (0); nextBtn = $ ("input # nextButton")। (0); $ (पेरेंट कैट्स) .change (पेरेंट चाइंगेड); $ (सब कैट) .change (चाइल्ड चेंंकेड);}); & Lt; / स्क्रिप्ट & gt; & Lt; / head & gt;

तब मैंने अपनी सभी इनलाइन स्क्रिप्ट को एक फ़ाइल (myScript.js) में रख दिया और मेरी एचटीएमएल इस पर बदल दिया: 

  & lt; head & gt; & Lt; script src = "/ स्क्रिप्ट / jquery-1.3.2.min.js" प्रकार = "टेक्स्ट / जावास्क्रिप्ट" & gt; & lt; / स्क्रिप्ट & gt; & Lt; script src = "/ स्क्रिप्ट / myScript.js" प्रकार = "टेक्स्ट / जावास्क्रिप्ट" & gt; & lt; / स्क्रिप्ट & gt; & Lt; / head & gt;

और अब कुछ भी काम नहीं कर रहा है मैंने अपना पृष्ठ आईई 7 में खोला और उसमें एक पृष्ठ त्रुटि आई:

रेखा: 54
त्रुटि: अज्ञात नाम।

रेखा 54 मेरी बाहरी जावास्क्रिप्ट फाइल की अंतिम पंक्ति होती है।

मैं क्या कर रहा हूं?

क्या मैं कह रहा हूं कि यह एएसपी.नेट है? यदि ऐसा है, इनलाइन स्क्रिप्ट की तरह: 

  & lt;% = Url.Action ("GetDescription")% & gt;

बाहरी जावास्क्रिप्ट फ़ाइल में नहीं जा सकता।


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