PHP:无法在echo中调用onclick函数


PHP: Cant call a onclick function inside an echo

我该怎么做才能调用onclick函数??真的需要帮助,事先谢谢:)

  if ($result1 = mysqli_query($con, $SQL1)){
    while ($row1=mysqli_fetch_array($result1)) {
      echo "
    <h1>$row1[Navn]</h1>
    <hr>
    <div id='Billedinfo'>
      <div id='Billed'>
        <div id='Hoved'>
          <img id='billedet' src='Billeder/$row1[Billed]'>
        </div>
        <div id='Billederne'>
          <img class='Andrebilleder' id='photo1' src='Billeder/$row1[billed1]'onmouseover='hover('photo1')' onmouseout='hoverout('photo1')' onclick='chooser('photo1')'>
          <img class='Andrebilleder' id='photo2' src='Billeder/$row1[billed2]' onmouseover='hover('photo2')' onmouseout='hoverout('photo2')' onclick='chooser('photo2')'>
          <img class='Andrebilleder' id='photo3' src='Billeder/$row1[billed3]' onmouseover='hover('photo3')' onmouseout='hoverout('photo3')' onclick='chooser('photo3')'>
          <img class='Andrebilleder' id='photo4' src='Billeder/$row1[billed4]' onmouseover='hover('photo4')' onmouseout='hoverout('photo4')' onclick='chooser('photo4')'>
          <img class='Andrebilleder' id='photo5' src='Billeder/$row1[billed5]' onmouseover='hover('photo5')' onmouseout='hoverout('photo5')' onclick='chooser('photo5')'>
          <img class='Andrebilleder' id='photo6' src='Billeder/$row1[billed6]' onmouseover='hover('photo6')' onmouseout='hoverout('photo6')' onclick='chooser('photo6')'>
          <img class='Andrebilleder' id='photo7' src='Billeder/$row1[billed7]' onmouseover='hover('photo7')' onmouseout='hoverout('photo7')' onclick='chooser('photo7')'>
          <img class='Andrebilleder' id='photo8' src='Billeder/$row1[billed8]' onmouseover='hover('photo8')' onmouseout='hoverout('photo8')' onclick='chooser('photo8')'>
        </div>

此外,您在插入php变量时也有错误。

将$row1[Navn]替换为$row1['Navn'],所有php变量也是如此。

并像这样调用js函数,

 onmouseout='"hoverout('photo8')'"

您应该查看生成的HTML,因为您在没有任何考虑的情况下将引号包含在其他引号中,所以它非常混乱。

要将一个引号包含在同一个引号的另一个集合中,需要使用'对其进行转义。

onmouseover='hover('photo6')'将变为onmouseover='hover(''photo6'')'

考虑到将数组索引引用为常量而不是字符串($row1[billed8]应该是$row1['billed8']),您还应该收到PHP警告。