Question

有人可以帮助我吗，我有一个PHP搜索脚本（如下所示，周围有HTML），我想在必要时返回多个搜索结果，但目前还没有，我已经挣得足够了到目前为止，我迷失了......

<?php
                    if(isset($_POST['completedsearch']))
                    {
                            $term = $_POST['query'];
                            $mysql = mysql_connect("localhost","user","password");
                            mysql_select_db("hcsd");
                            $qu = mysql_query("SELECT * FROM Sheet1 WHERE COMPANY LIKE '%".mysql_real_escape_string($term)."%' OR LOCATION LIKE '%"$.mysql_real_escape_string($term)."%' OR KEYWORDS LIKE '%".mysql_real_escape_string($term)."%' OR PRODUCTSSERVICES LIKE '%".mysql_real_escape_string($term)."%' "); //selects the row that contains ANYTHING like the submitted string
                            echo "
                                            <table><tr><th>Name</th>
                                            <th>Location</th>
                                            <th>Products/Services</th></tr>
                                            ";
                            while($row = mysql_fetch_array($qu))
                                       {

                                            echo "<tr><td>";  
                                            echo $row['COMPANY'];
                                            echo "</td>";
                                            echo "<td>";
                                            echo $row['LOCATION'];
                                            echo "</td>";
                                            echo "<td>";
                                            echo $row['PRODUCTSSERVICES'];
                                            echo "</td></tr></table>";
                            }
                    }
            ?>

通过搜索电视可以在www.healthcareservicesdirect.com上查看搜索，该电视应显示Sky＆amp;处女，但截至目前，只显示天空

Answer 1

尝试

"SELECT * FROM Sheet1 WHERE COMPANY LIKE '%".mysql_real_escape_string($term)."%' OR LOCATION LIKE '%".mysql_real_escape_string($term)."%' OR KEYWORDS LIKE '%".mysql_real_escape_string($term)."%' OR PRODUCTSSERVICES LIKE '%".mysql_real_escape_string($term)."%'"

我注意到LOCATION LIKE '%"$.mysql_real_escape_string($term)."%'你有$

Answer 2

我刚刚看了它，我看到它列出处女！它只是不在表中，因为第一个条目是。它位于页面底部。要解决这个问题，你必须将echo“”从while循环中删除，因为在第一次输入后你将关闭表格，所以你的代码应该是这样的：

<?php
                if(isset($_POST['completedsearch']))
                {
                        $term = $_POST['query'];
                        $mysql = mysql_connect("localhost","user","password");
                        mysql_select_db("hcsd");
                        $qu = mysql_query("SELECT * FROM Sheet1 WHERE COMPANY LIKE '%".mysql_real_escape_string($term)."%' OR LOCATION LIKE '%"$.mysql_real_escape_string($term)."%' OR KEYWORDS LIKE '%".mysql_real_escape_string($term)."%' OR PRODUCTSSERVICES LIKE '%".mysql_real_escape_string($term)."%' "); //selects the row that contains ANYTHING like the submitted string
                        echo "
                                        <table><tr><th>Name</th>
                                        <th>Location</th>
                                        <th>Products/Services</th></tr>
                                        ";
                        while($row = mysql_fetch_array($qu))
                                   {

                                        echo "<tr><td>";  
                                        echo $row['COMPANY'];
                                        echo "</td>";
                                        echo "<td>";
                                        echo $row['LOCATION'];
                                        echo "</td>";
                                        echo "<td>";
                                        echo $row['PRODUCTSSERVICES'];
                                        echo "</td></tr>";
                        }
                        echo "</table>";
                }
        ?>

Answer 3

echo "
<table><tr><th>Name</th>
<th>Location</th>
<th>Products/Services</th></tr>
";
while($row = mysql_fetch_array($qu))
{
    echo "<tr><td>";  
    echo $row['COMPANY'];
    echo "</td>";
    echo "<td>";
    echo $row['LOCATION'];
    echo "</td>";
    echo "<td>";
    echo $row['PRODUCTSSERVICES'];
    echo "</td></tr>";
}
echo "</table>";

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PHP搜索只显示一个结果？

3 个答案: