Question

我编写了一个PHP脚本，用于从SQL数据库中提取数据以显示在网页上。一切似乎都很好但是当我运行脚本时它会抛出错误：

解析错误：语法错误，意外'）'，期待';'在第21行的db_data.php中

第21行是我的FOR循环，我检查了循环的语法，它似乎是正确的，所以我无法理解为什么它失败了。

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

echo "<table border='1'>
<tr>
<th>Ref</th>
<th>Nameame</th>
<th>Location</th>
<th>Email</th>
<th>Issue</th>
<th>Urgency</th>
</tr>";

for($row = mysqli_fetch_array($result)){
    echo "<tr>";
    echo "<td>" . $row['REF'] . "</td>";
    echo "<td>" . $row['NAME'] . "</td>";
    echo "<td>" . $row['LOCATION'] . "</td>";
    echo "<td>" . $row['EMAIL'] . "</td>";
    echo "<td>" . $row['ISSUE'] . "</td>";
    echo "<td>" . $row['URGENCY'] . "</td>";
    echo "</tr>";
}
echo "</table>";

mysqli_close($con);

Answer 1

更改为

<?php
//                   Missing FROM here vv
$result = mysqli_query($con,"SELECT * FROM igi");

echo "<table border='1'>
<tr>
<th>Ref</th>
<th>Nameame</th>
<th>Location</th>
<th>Email</th>
<th>Issue</th>
<th>Urgency</th>
</tr>";

while($row = mysqli_fetch_array($result)){
   echo "<tr>";
   echo "<td>" . $row['REF'] . "</td>";
   echo "<td>" . $row['NAME'] . "</td>";
   echo "<td>" . $row['LOCATION'] . "</td>";
   echo "<td>" . $row['EMAIL'] . "</td>";
   echo "<td>" . $row['ISSUE'] . "</td>";
   echo "<td>" . $row['URGENCY'] . "</td>";
   echo "</tr>";
}
echo "</table>";

mysqli_close($con);

Answer 2

正如Anthony Thompson和RJParick所说，你必须将for更改为while。您遇到的错误归因于for，需要3个这样的参数：

for($i; $i<10; $i++) {//Note the ; used to separate params.
    //do something
}

所以你的循环会像这样开始：

while($row = mysqli_fetch_array($result)){

另一个细节，你的SQL不正确使用SELECT * FROM igi。

For循环在PHP脚本中不起作用

2 个答案: