Question

我想在PHP中运行javascript，但这段代码对我来说根本不起作用。

if(!mysqli_fetch_assoc($result))
    {
      echo '<script type="text/javascript">',
     'document.getElementById("console1").innerHTML += "Query did not work";',
     '</script>'  ;

    }

Answer 1

将,替换为.以链接字符串。

if(!mysqli_fetch_assoc($result))
    {
      echo '<script type="text/javascript">' . 
      'document.getElementById("console1").innerHTML += "Query did not work";' .
      '</script>';
    }

Answer 2

您可以尝试Heredoc表示法：

<?php
    echo<<<HTML
    .
    .
    <script type="text/javascript">
      document.getElementById("console1").innerHTML += "Query did not work";
    </script>
    HTML; 

//[*Note that echo<<<HTML and HTML; has to be on the very left side without blank spaces]

Answer 3

你不能使用＆＃34;在PHP中，这对我有用！

echo "<script> 
       document.getElementById('console1').innerHTML += 'Query did not work'; 
      </script>";

Answer 4

因为它是用ajax完成的，所以尝试类似的东西：

<?php
    if(!mysqli_fetch_assoc($result)) {
        http_response_code(500);
        die(['error' => 'Query did not work!']);
    }
?>

并在你的前端代码中：

<script>
    $.get('your/query/path?query=...', function() {
        console.log('executed');
    }).fail(function(result) {
        $('#console1').append(result.error);
    });
</script>

如何在PHP中使用document.getElementbyId

4 个答案: