我的数据库搜索表单存在问题。该页面大部分运行良好,但我得到两个错误。
警告:ociexecute()[function.ociexecute]:ORA-00904:* $ search_term *:第53行 * * / search_results.php中的标识符无效
警告:ocifetch()[function.ocifetch]:ORA-24374:在获取或执行之前定义未完成,并在第70行上的** / search_results.php中获取
我评论了相关的行号。如果有人能告诉我为什么会这样,我会非常感激:)
编辑:忘记提及,尽管输入了应该得到结果的搜索字词,但表格中没有显示任何内容
编辑2:当查询更改为SELECT * FROM Phones时,它可以正常工作。它必须与查询有关。
<?php
$search_term = $_POST['search_box'];
/* Set oracle user login and password info */
$dbuser = "**"; /* your login */
$dbpass = "**"; /* your oracle access password */
$db = "**";
$connect = OCILogon($dbuser, $dbpass, $db);
if (!$connect) {
echo "An error occurred connecting to the database";
exit;
}
/* build sql statement using form data */
$query = "SELECT * FROM Phones WHERE Name LIKE ".$search_term;
/* check the sql statement for errors and if errors report them */
$stmt = OCIParse($connect, $query);
echo "SQL: $query<br>";
if(!$stmt) {
echo "An error occurred in parsing the sql string.\n";
exit;
}
OCIExecute($stmt); //line 53
?>
<h1 class="green">PHP and Oracle databases</h1>
<h4>Table: <em>Phones</em></h4>
<div align="center">
<table width="850" border="0" bgcolor="#339933" cellpadding="5" cellspacing="1">
<tr bgcolor="#006633">
<td width="75">ID</td>
<td width="75">Name</td>
<td width="100">Brand</td>
<td width="75">Photo</td>
</tr>
<?php
// Display all the values in the retrieved records, one record per row, in a loop
while(OCIFetch($stmt)) { //line 70
// Start a row for each record
echo("<tr valign=top bgcolor=#ccffcc>");
// Output fields, one per column
// Drainage value in column one
$fg1 = OCIResult($stmt,"ID"); //"ID number";
echo("<td width=75>");
echo ($fg1);
echo("</td>");
// Aspect value in column two
$fg2 = OCIResult($stmt,"NAME");//"Name of product";
echo("<td width=75>");
echo ($fg2);
echo("</td>");
// Temperature value in column three
$fg3 = OCIResult($stmt,"BRAND");//"Brand of product";
echo("<td width=75>");
echo ($fg3);
echo("</td>");
// Height value in column four
$fg4 = OCIResult($stmt,"PHOTO");//"Photo file path";
echo("<td width=75>");
echo ($fg4);
echo("</td>");
// End the row
echo("</tr>");
}
// Close the connection
OCILogOff ($connect);
?>
答案 0 :(得分:1)
看起来您的SQL命令需要围绕$search_term
内容的引号。
我不懂PHP,但我想它应该是这样的:
$query = "SELECT * FROM Phones WHERE Name LIKE '%".$search_term."%'";
请注意添加撇号(SQL的引号字符)和百分号(LIKE的通配符)。我不确定你在PHP中如何正确引用撇号(')和百分号(%),所以你应该检查一下。
(抱歉,我不知道如何在Oracle上进行不区分大小写的搜索,你应该将其作为一个单独的问题发布。)