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我无法使 MySQL Source 值等于 $source: $result = mysql_query("SELECT Source FROM videos WHERE Name='$name'"); $source=['Source'];

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2 回答 2

0
$result = mysql_query("SELECT Source FROM videos WHERE Name='$name'"); 
$ds = mysql_fetch_object($result);
$source= $ds -> Source;

这可以做到,但你应该使用 mysqli

于 2013-02-08T13:39:10.977 回答
0
$sql = "SELECT Source FROM videos WHERE Name = '$name'";
$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);

$your_var = $row['Source'];
于 2013-02-08T13:45:19.207 回答