小编use*_*936的帖子

我有一个SQLite数据库,我正在尝试使用PHP连接它.这就是我正在使用的:

<?php
    $dbconn = sqlite_open('combadd.sqlite');

    if ($dbconn) {
        $result = sqlite_query($dbconn,  "SELECT * FROM combo_calcs WHERE options='easy'");
        var_dump(sqlite_fetch_array($result, SQLITE_ASSOC));
    } else {
        print "Connection to database failed!\n";
    }
?>

但是,我收到此错误:

警告:sqlite_open()[function.sqlite-open]:文件已加密或不是C:\xampp\htdocs\deepthi\combadd\combadd_db.php第4行的数据库
连接数据库失败!

怎么了,怎么解决？

单击单选按钮时,我有以下代码是,如果不是必须禁用,则必须启用文本框.请有人帮助我.谢谢

<div class="label_left">
    <label>Does the tourism office operate on a biennial budget? : (Y/N)</label>
</div>
<div class="text_right">
    <br>
    <br>
    <input type="radio" id="high" name="high" />
    <label>Y</label>&nbsp;&nbsp;&nbsp;&nbsp;
    <input type="radio" id="high" name="high" />
    <label>N</label>
    <br />
    <br />
    <br />
</div>
<div class="label_left">
    <label>If YES,please indicate when the current biennial budget began: (mm/dd/yy)</label>
</div>
<div class="text_right">
    <br>
    <br>
    <input type="text" name="date_biennial_budget" id="date_biennial_budget" value="" size="30" />
    <br />
    <br />
    <br />
</div>