Javascript 在新窗口中打开图像的脚本不工作

Javascript 在新窗口中打开图像的脚本不工作,javascript,php,Javascript,Php,我正在使用此代码在新窗口中打开图像,但无法使其正常工作。怎么了 echo " <td align='center'> <a href='images/spasergjengen/" . $row['Grad'] . "' target='_blank'> <img src='images/spasergjengen/" . $row['Grad'] . "' width='125' height='150'

我正在使用此代码在新窗口中打开图像,但无法使其正常工作。怎么了

echo "
<td align='center'>
<a href='images/spasergjengen/" . $row['Grad'] . "'  target='_blank'>
  <img src='images/spasergjengen/" . $row['Grad'] . "' width='125' height='150' 
  title='Spasergjengen' alt='Spasergjengen' />
</a>
<br>
<button onclick='myFunction" . $row['MedlemId'] . "'()'>Se Bilde</button>
<script>
   function myFunction" . $row['MedlemId'] . "'() {     
   window.open('images/spasergjengen/" . $row['Grad'] . "', '_blank',
  'toolbar=yes,scrollbars=yes,resizable=yes
  ,top=200,left=300,width=750,height=565'); } 
</script>
</td>
";
echo”

[测试]

<?php  echo "<td align='center'><a href='images/spasergjengen/" . $row['Grad'] . "'  target='_blank'><img src='images/spasergjengen/" . $row['Grad'] . "' width='125' height='150' title='Spasergjengen' alt='Spasergjengen' /></a><br><button onclick='myFunction" . $row['MedlemId'] . "()'>Se Bilde</button>
<script> function myFunction" . $row['MedlemId'] . "() {     window.open('images/spasergjengen/" . $row['Grad'] . ",_blank', 'toolbar=yes,scrollbars=yes,resizable=yes,top=200,left=300,width=750,height=565'); } </script></td>";
?>

Typo:在第一行,你有一个额外的引号:
“()”
应该是
“()”
,在第二行:
“()”
应该是
”()
,如果你在浏览器中使用
查看源代码,你应该会看到这些。你也应该在JavaScript控制台中看到语法错误。