Question

所以这是我的代码，但它不起作用，我只是得到了这个

Title Article "; while($row = mysqli_fetch_array($result)) { echo ""; echo "" . $row['Title'] . ""; echo "" . $row['Article'] . ""; echo ""; } echo ""; mysqli_close($con); ?>

有什么想法吗？

<?php
$con=mysqli_connect("localhost","____","____","test_database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM written");

echo "<table border='1'>
<tr>
<th>Title</th>
<th>Article</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['Title'] . "</td>";
  echo "<td>" . $row['Article'] . "</td>";
  echo "</tr>";
  }
 echo "</table>";

 mysqli_close($con);
 ?>

Answer 1

Tripple检查您的引号是否设置正确。任

echo "<table border='1'>

或

</tr>";

有错误。

您也可以尝试使用echo作为函数（手镯），它应该更容易发现错误： - ）

查看生成的标记而不是文本会很有帮助; - ）

Answer 2

你必须这样做：

echo "<table border='1'><tr><th>Title</th><th>Article</th></tr>";

或

echo "<table border='1'>";
echo "<tr>";
echo "<th>Title</th>";
echo "<th>Article</th>";
echo "</tr>";

而不是

echo "<table border='1'>
<tr>
<th>Title</th>
<th>Article</th>
</tr>";

检索Mysql数据失败了......为什么？

2 个答案: