asp.net - ReportViewer and font-family -

-

i implemented web-portal in .net integrate reportviewer control load ssrs reports. in order customize reports use own font in rdl report file. font dynamically loaded on client side font-family declaration in css it:

@font-face{ font-family: 'helvetica 35 thin'; src:url('../fonts/helvneue35_w1g.eot?#iefix'); src:url('../fonts/helvneue35_w1g.eot?#iefix') format('eot'), url('../fonts/helvneue35_w1g.woff2') format('woff2'), url('../fonts/helvneue35_w1g.woff') format('woff'), url('../fonts/helvneue35_w1g.ttf') format('truetype'), url('../fonts/helvneue35_w1g.svg') format('svg');

}

all work perfectily on localy visual studio iis express : font it's correctly loaded on each web browser , correctly displayed in report. works on development server. sincei deployed on integration server, font not displayed. specify use same ssrs server on local visual , on integration platform.

i investigated, , noticed difference between ssrs code generated reportviewer control iis express , iis 8.

in fact visual studio execution found :

    .a6b8151803b1943dc956fcd4b577a57f413{word-wrap:break-word;     white-space:pre-wrap;padding-left:2pt;padding-top:2pt;     padding-right:2pt;padding-bottom:2pt;border:1pt none black;     background-color:transparent;font-style:normal;     font-family:'helvetica 35 thin';

and iis 8 found :

    .a87918d1b077d4156abf69d70d7bd400213{word-wrap:break-word;     white-space:pre-wrap;padding-left:2pt;padding-top:2pt;     padding-right:2pt;padding-bottom:2pt;border:1pt none black;     background-color:transparent;font-style:normal;     font-family:helvetica 35 thin;

the quotes have been removed iis 8. font name contains space character, explain why font not found browser.

now need explain why quote removed platform. have encountered issue?

thank help.

we've experienced issue. fixed using expression font name , setting

="'font space in name'"

if using ie using compatibility mode may fix issue you.


Comments

Post a Comment

Popular posts from this blog

Hatching array of circles in AutoCAD using c# -

-
i beginner in autocad plugins .i have created array of circle using following code. now, have hatch circles in array .how that??how hatch circles @ once? moreover, how can return hatch method , use hatch objects.sorry language.thanks in advance. [assembly: commandclass(typeof(beamsection.class1))] namespace beamsection { public class class1 { //class1 obj = new class1(); [commandmethod("beamatsupport")] public void addlightweightpolyline() { // current document , database document acdoc = application.documentmanager.mdiactivedocument; database accurdb = acdoc.database; promptpointresult ppr = acdoc.editor.getpoint("\nselect starting point "); var ucs = acdoc.editor.currentusercoordinatesystem; point3d startingpt = ppr.value.transformby(ucs); list<double> radius = new list<double>(); radius.add(0.5);
Read more

ios - UITEXTFIELD InputView Uipicker not working in swift -

-
hello trying pop uipickerview view programmatically when user clicks on textfield. have tried doesn't doing anything. isn't working. nothing happening when click textfield class userprofiletableviewcontroller: uitableviewcontroller,uitextfielddelegate,uipickerviewdelegate,uipickerviewdatasource { var itempicker: uipickerview! = uipickerview() @iboutlet weak var gendertxtfield: uitextfield! var gender = ["male","female"] override func viewdidload() { super.viewdidload() gendertxtfield.delegate = self itempicker!.delegate = self itempicker!.datasource = self itempicker!.backgroundcolor = uicolor.blackcolor() self.gendertxtfield.inputview = itempicker } func numberofcomponentsinpickerview(pickerview: uipickerview) -> int{ return 1 } // returns # of rows in each component.. func pickerview(pickerview: uipickerview, numberofrowsincomponent component:
Read more

Python Pig Latin Translator -

-
so have been working on @ codecademy.com while trying learn python. don't know if blind , can't see why won't work or what. here's code won't let me use. function print variable if had typed in , print "empty" if didn't type anything. print "welcome english pig latin translator!" original = raw_input("what's name?") if len(original) > 0 print original else print "empty" if len(original) > 0 else the above lines contains syntax errors. try: if len(original) > 0: else: note added colons, ( : ).
Read more