jquery - javascript: string.replace doesn't work -

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इस सवाल का पहले से ही एक उत्तर है: < / P>

  • 34 उत्तर

    मेरे साथ थोड़ा समस्या है जावास्क्रिप्ट / jquery में एक स्ट्रिंग की जगह सबसे पहले कोड:

    HTML: 

      & lt; fieldet & gt; & Lt; किंवदंती & gt; संपर्क व्यक्ति & lt; / किंवदंती & gt; & Lt; div id = "cm_contactPersons" & gt; & Lt; fieldet id = "cm_% ईमेल%" & gt; & Lt; किंवदंती & gt; & lt; span class = "glyphicon glyphicon-circle-arrow-right" & gt; & lt; / span & gt; % ईमेल% & lt; / कथा & gt; & Lt; div वर्ग = "फॉर्म-समूह" & gt; & Lt; label = "cm_% email_inputFirstname" class = "col-sm-2 control-label" & gt; फर्स्टनाम & lt; / label & gt; & Lt; div class = "col-sm-10" & gt; & Lt; input type = "text" class = "form-control" id = "cm_% email% _inputFirstname" मान = "% firstname%" प्लेसहोल्डर = "" आवश्यक & gt; & Lt; / div & gt; & Lt; / div & gt; & Lt; div वर्ग = "फॉर्म-समूह" & gt; & Lt; label = "cm_% email% _inputLastname" वर्ग = "col-sm-2-control-label" & gt; अंतिम नाम & lt; / label & gt; & Lt; div class = "col-sm-10" & gt; & Lt; input type = "text" class = "form-control" id = "cm_% ईमेल% _inputLastname" value = "% lastname%" प्लेसहोल्डर = "i। अधिकतम" आवश्यक & gt; & Lt; / div & gt; & Lt; / div & gt; & Lt; div वर्ग = "फॉर्म-समूह" & gt; & Lt; label class = "col-sm-2-control-label" & gt; क्रियाएं & lt; / label & gt; & Lt; div class = "col-sm-10" & gt; & Lt; बटन प्रकार = "बटन" वर्ग = "बीटीएन बीटीएन-चेतावनी" & gt; नया पासवर्ड बनाएं & lt; / button & gt; & Lt; बटन प्रकार = "बटन" वर्ग = "बीटीएन बीटीएन-खतरे" & gt; संपर्क व्यक्ति हटाएं & lt; / button & gt; & Lt; / div & gt; & Lt; / div & gt; & Lt; / fieldset & gt; & Lt; / div & gt; & Lt; / fieldset & gt;   
     जावास्क्रिप्ट:।  
      $ (document) .ready (function () {$ ( '# cm_customerList') btsListFilter ( '# cm_customerSearchInput'); वर editingArea = $ ( '# cm_editingArea'); वर contactPersons = $ ( '# cm_contactPersons'); वर contactPersonsTpl = contactPersons.html (); वर alertArea = $ ( '# cm_alertArea'); $ ( '# cm_customerList & gt; ') .प्रत्येक (function () {$ (this) .click (समारोह (ई) {e.preventDefault (); $ (' # cm_customerList & gt; a.active '।) removeClass (' सक्रिय '); $ (this) । .addClass ( 'सक्रिय', 400); $ ( 'एचटीएमएल') चेतन ({scrollTop: 0}); alertArea.fadeOut (400, function () {alertArea.empty ();}); $ .ajax ({ प्रकार: 'प्राप्त करके' यूआरएल: + $ (this) .attr ( 'डेटा-ajaxUrl'), सफलता 'index.php?': समारोह (डेटा) {रेस = $ .parseJSON (डेटा); editingArea.fadeOut (400 , function () {$ ( '# cm_inputName') वैल (res.name);। $ ( '# cm_inputStreet') वैल (res.street);। $ ( '# cm_inputPostalcode') वैल (res.postalcode)। $ ('# Cm_inputCity')। Val (res.city); $ ('# सेमी_इनपुटदेश विकल्प [मान = "' + रिज्यूशन + '"]')। Prop ('selected', true); $ ('# cm_inputCoun कोशिश ')। चयन 2 ({width:' 100% '}); $ ('# सेमी_इनपुट ट्यूटर्स विकल्प: चयनित')। प्रोप ('चयनित', गलत); $ .each (res.tutors, कार्य (कुंजी, शिक्षक) {$ ('# cm_inputTutors विकल्प [value = "' + tutor.username + '"]')। प्रोप ('चयनित', सच);}); $ ('# सेमी_इनपुटटुटर्स')। चयन 2 ({चौड़ाई: '100%'}); contactPersons.empty (); $ .each (res.contactpeople, फ़ंक्शन (कुंजी, संपर्कप्रसारण) {contactPersons.append (संपर्क PersonsTpl.replace ('% ईमेल%', संपर्क प्रपत्र));}; editingArea.fadeIn (); }); }, त्रुटि: फ़ंक्शन (xhr, ajaxOptions, thrownError) {// त्रुटि हैंडलर}}); }); }); });   
     " contactPersons.append (contactPersonsTpl.replace ('% ईमेल%', संपर्क फ़ॉर्मेट) के साथ हिस्सा; " प्रतिस्थापन बस काम नहीं करता है मैंने सभी विभिन्न प्रकार के डीबगिंग (लॉग, उपस्ट्रॉफ्ट बदलते आदि) की कोशिश की। सवाल ही नहीं।  
     लेकिन अगर मैं लिखता हूं " $ ('# cm_contactPersons')। Html ()। Replace ('% email%', 'test@blubb.at'); " में फ़ायरफ़ॉक्स कंसोल, यह काम करता है  
     मैं क्या कर रहा हूँ गलत?   
     
     
     रेगेक्स का उपयोग करें ..  
     डिफ़ॉल्ट रूप से प्रतिस्थापित केवल पैटर्न की पहली घटना बदल जाएगी। जी रूपांतरण के साथ नियमित अभिव्यक्ति का उपयोग करने के लिए स्ट्रिंग में सभी पैटर्न को बदलने के लिए  <प्री>  contactPersons.append (संपर्क PersonsTpl.replace (/% ईमेल% / जी, संपर्क स्वरूप));

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