Question

我想检查数据是否为空。这是空的，即使不是谷歌链接，也不会显示任何内容。如果有数据，则显示谷歌链接。我该如何解决？

$result = $mysqli->query("SELECT * FROM teams WHERE teamid = ".$_GET['teamid']." ORDER BY `teamname` DESC");

   $teamdetails = mysqli_fetch_assoc($result);

echo '<table id=kalender_table><tr><td><h3>'.$teamdetails['teamname'].'</h3> <br>';
echo ''.$teamdetails['teamid'].'<br>';
echo '<a href="'.$teamdetails['website'].'" target="_blank">'.$teamdetails['website'].'</a> <br></td>';
echo '<td><img src=../../logo/'.$teamdetails['image'].'></td></tr>';
echo '<tr><td colspan="2">'.$teamdetails['cmp1_name'].'</td></tr>';
echo '<tr><td colspan="2">'.$teamdetails['cmp1_adress'].'</td></tr>';
echo '<tr><td colspan="2">'.$teamdetails['cmp1_zip'].' '.$teamdetails['cmp1_city'].'</td></tr>';
echo '<tr><td colspan="2">'.$teamdetails['cmp1_phone'].'</td></tr>';
echo '<tr><td colspan="2"><a href="https://www.google.be/maps/place/'.$teamdetails['cmp1_adress'].'+'.$teamdetails['cmp1_zip'].'+'.$teamdetails['cmp1_city'].'" target="_blank">Google maps</a></td></tr>';

Answer 1

你可以试试这个：

<?php
if (is_array($teamdetails) && count($teamdetails) > 0) {
    // Do something
}

Answer 2

简单地使用行计数，如果大于0则显示，如果没有，则无需显示

$result = $mysqli->query("SELECT * FROM teams WHERE teamid = ".$_GET['teamid']." ORDER BY `teamname` DESC");
$teamdetails = mysqli_fetch_assoc($result);

if((mysqli_num_rows($result) > 0) {
    echo '<table id=kalender_table><tr><td><h3>'.$teamdetails['teamname'].'</h3> <br>';
    echo ''.$teamdetails['teamid'].'<br>';
    echo '<a href="'.$teamdetails['website'].'" target="_blank">'.$teamdetails['website'].'</a> <br></td>';
    echo '<td><img src=../../logo/'.$teamdetails['image'].'></td></tr>';
    echo '<tr><td colspan="2">'.$teamdetails['cmp1_name'].'</td></tr>';
    echo '<tr><td colspan="2">'.$teamdetails['cmp1_adress'].'</td></tr>';
    echo '<tr><td colspan="2">'.$teamdetails['cmp1_zip'].' '.$teamdetails['cmp1_city'].'</td></tr>';
    echo '<tr><td colspan="2">'.$teamdetails['cmp1_phone'].'</td></tr>';
    echo '<tr><td colspan="2"><a href="https://www.google.be/maps/place/'.$teamdetails['cmp1_adress'].'+'.$teamdetails['cmp1_zip'].'+'.$teamdetails['cmp1_city'].'" target="_blank">Google maps</a></td></tr>';
} else {
    echo '<tr><td colspan="2">'Nothing to Show'</td></tr>';
}

Answer 3

运行mysqli_fetch_assoc之后

：

if (mysqli_affected_rows()){
...
}

Answer 4

您需要使用“isset”函数，因为使用count（）可能存在行但其中没有数据。

[...]
$teamdetails = mysqli_fetch_assoc($result);

if (isset($teamdetails['teamname']) & $teamdetails['teamname']) {
  echo '<table id=kalender_table><tr><td><h3>'.$teamdetails['teamname'].'</h3> <br>';
  [...]
}
[...]

如何检查有数据，然后什么都不显示

4 个答案: