我想知道这段代码是否可以更好地用 jQuery 编写。该代码在 Firefox 中运行良好,但在 safari 或 chrome 中无法运行,并且没有给出任何错误,所以我很难弄清楚它为什么不起作用。
任何帮助或想法表示赞赏...
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<area onmouseout="if(document.images) document.getElementById('Image-Maps_8201304160533561').src= 'http://www.chrisjonesortho.com.au/wp-content/themes/medicate/images/Human-Image_Chris-Jones.png';" onmouseover="if(document.images) document.getElementById('Image-Maps_8201304160533561').src='http://www.chrisjonesortho.com.au/wp-content/themes/medicate/images/knee.png';" title="" alt="" href="http://www.chrisjonesortho.com.au/patient-information/" coords="0,99,90,137" shape="rect">
<area onmouseout="if(document.images) document.getElementById('Image-Maps_8201304160533561').src= 'http://www.chrisjonesortho.com.au/wp-content/themes/medicate/images/Human-Image_Chris-Jones.png';" onmouseover="if(document.images) document.getElementById('Image-Maps_8201304160533561').src= 'http://www.chrisjonesortho.com.au/wp-content/themes/medicate/images/foot.png';" title="" alt="" href="http://www.chrisjonesortho.com.au/patient-information/" coords="0,197,90,235" shape="rect">