0

所以这是我的代码,但它不起作用,我只是得到这个

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

有任何想法吗?

<?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);
 ?>
4

2 回答 2

1

三重检查您的引号设置是否正确。任何一个

echo "<table border='1'>

或者

</tr>";

有一个错误。

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;-)

于 2013-11-04T18:21:10.723 回答
0

你必须这样做:

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

或者

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

代替

echo "<table border='1'>
<tr>
<th>Title</th>
<th>Article</th>
</tr>";
于 2013-11-05T13:54:45.387 回答