Resolution check in illustrator

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How to check the raster image resolution in illustrator through script?

Scripting

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M

moluappleCorrect answer

Inspiring

This seems to solve it:

// select a rasterItem, no rotation or skew
72/app.selection[0].matrix.mValueA;
// also
72/app.selection[0].matrix.mValueD;
// for rotation applied(no skew) item
var item = app.selection[0],
    matrix = item.matrix,
    rotation = 180/Math.PI * Math.atan2(matrix.mValueC, matrix.mValueD);
item.rotate(rotation);
alert(72/item.matrix.mValueA);
item.rotate(-rotation);

Stephen Marsh

Community Expert

As a scripting newb, the following is beyond me, however I would do something like:

Embedded image pixel width, divided by physical width of image on canvas in inches = width ppi value

Embedded image pixel height, divided by physical height of image on canvas in inches = height ppi value

Pick the smallest value of the two (they should be close), one may need to round the value to the closest integer

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wckdtall

Inspiring

Illustrator makes this difficult, and not straight forward... plus percentage math isn't my forte. One can read the percentage of the image from the matrix mValueB and C, but if it's Linked the percentage is based on the resolution of the image. So as an example, we get:

LINKED: W: 1800px | W": 25in" and Link Panel reads W: 3300px | Scale: 113% | Resolution: 132dpi

EMBEDDED: W: 1800px | W": 25in" and Link Panel reads W: 3300px | Scale: 54.6% | Resolution 132dpi

The source image in photoshop reads 150dpi, but being that DPI is not readable in AI, the image must be embedded and at 0 rotation and then one can find the original image size and DPI by doing some math.

(Wpx / Scale) * 100 = Original Wpx
(Original Wpx / Current Wpx) * 72 = DPI of Image.

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wckdtall

Inspiring

Now I see the beauty of Moluapple's first block of code. If someone has a quick breakdown as to why DPI/Scale % = Resolution, I'd appreciate it. Also, There doesn't seem to be a clear explanation in docs on what the matrix is comprised of. A and D, seem to be the decimal form of scale %, and I'd assume TX and TY are Transformation? That leaves mValueB and mValueC.

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M

moluappleCorrect answer

Inspiring

This seems to solve it:

// select a rasterItem, no rotation or skew
72/app.selection[0].matrix.mValueA;
// also
72/app.selection[0].matrix.mValueD;
// for rotation applied(no skew) item
var item = app.selection[0],
    matrix = item.matrix,
    rotation = 180/Math.PI * Math.atan2(matrix.mValueC, matrix.mValueD);
item.rotate(rotation);
alert(72/item.matrix.mValueA);
item.rotate(-rotation);

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