PHP代码未执行

Question

$sql1=mysql_query("SELECT * FROM Persons", $con);

echo "<table border="3">
        <tr>
        <th>Name</th>
        <th>Age</th>
        </tr>";
while($info=mysql_fetch_array($sql1))
 {
        echo "<tr>";
        echo "<td>" . $info['fname'] . "</td>";
        echo "<td>" . $info['age'] . "</td>";
        echo "</tr>";
 }
echo "</table>";

this code is a part of a code which is trying to retrieve data from the table"Persons", there is some error in this part of the code..

Answer 1

You have double quotes in the quoted html. Try using single quotes in stead, ie

echo "<table border='3'> <--- here
        <tr>
        <th>Name</th>
        <th>Age</th>
        </tr>";

Answer 2

Your code looks ok, except for the unescaped double quotes:

It should be:

echo "<table border=\"3\"> ... ";

or

echo '<table border="3"> ... ';

Make sure it is enclosed in <?php and ?> .

Also make sure your db columns names of fname and age really exist....

Make sure you're getting what you think back from the DB, by using print_r($info) or var_dump($info) .

Finally, your connection $con could be broken / not working. You can check that by using:

if ( ! $con ) {
    die('Could not connect: ' . mysql_error());
}

$sql1 = mysql_query("SELECT * FROM Persons", $con);
...

Answer 3

Beside double quotes in second line.
mysql_fetch_array return array indexed by integer if you want asoc array indexed by fields use mysql_fetch_assoc

while ($info=mysql_fetch_assoc($sql1)) {
   ...
}