Php 从mysql数据库unicode进行Ajax搜索失败
我希望版主不要将这个问题标记为重复的问题,因为我对这个问题进行了将近1周的研究,无法找出问题所在 我有这段代码,用于从MySQL数据库表中搜索数据,它不读取Unicode字符。我的猜测是$\u GET function string读取Unicode字符,因此,我将MySQL表字符集设置为utf-8,列排序规则为utf8\u general\u ci,并将这两行添加到php代码中:Php 从mysql数据库unicode进行Ajax搜索失败,php,mysql,unicode,get,Php,Mysql,Unicode,Get,我希望版主不要将这个问题标记为重复的问题,因为我对这个问题进行了将近1周的研究,无法找出问题所在 我有这段代码,用于从MySQL数据库表中搜索数据,它不读取Unicode字符。我的猜测是$\u GET function string读取Unicode字符,因此,我将MySQL表字符集设置为utf-8,列排序规则为utf8\u general\u ci,并将这两行添加到php代码中: header('Content-Type: text/html; charset=utf-8'); mysqli_
header('Content-Type: text/html; charset=utf-8');
mysqli_set_charset($connect,'utf8');
但Unicode字符的结果仍然是0。如果我在输入中键入“a”,它会显示以“a”或“ä,åá…”开头的所有单词,但是,如果我键入一些Unicode字符,例如“ä”,它会显示“未找到结果”
代码也在线atm这里是链接
我还将utf-8元数据添加到HTML文件中。我希望有人告诉我怎么了。守则本身是:
header('Content-Type: text/html; charset=utf-8');
mysqli_set_charset($connect,'utf8');
if(isset($_GET['p'])) {
$page_number = $_GET['p'];
$arraySearch = $_GET['terms'];
$show_count = $_GET['count'];
settype($page_number, 'integer');
}
$nospaces = substr($_GET['terms'],1,4);
$offset = ($page_number - 1) * $records_number;
// check for an empty string and display a message.
if ($_GET['terms'] == "") {
echo '<div id="counter">Type "äää" or "ääää"!</div>';
// minim 3 characters condition
} else if(strlen($_GET['terms']) < $limitchar) {
echo '<div id="counter">'. $limitchar .' characters minimum</div>';
} else {
// explode search words into an array
$arraySearch = explode(" ", $_GET['terms']);
// table fields to search
$arrayFields = array(0 => $first_field, 1 => $second_field);
$countSearch = count($arraySearch);
$a = 0;
$b = 0;
$query = "SELECT * FROM $table_name WHERE (";
$countFields = count($arrayFields);
while ($a < $countFields)
{
while ($b < $countSearch)
{
$query = $query."$arrayFields[$a] LIKE '$arraySearch[$b]%'";
$b++;
if ($b < $countSearch)
{
$query = $query." AND ";
}
}
$b = 0;
$a++;
if ($a < $countFields)
{
$query = $query.") OR (";
}
}
$query = $query.") LIMIT $offset, $records_number;";
$search = mysqli_query($connect, $query);
// get number of search results
$arrayFields = array(0 => $first_field);
$countSearch = count($arraySearch);
$a = 0;
$b = 0;
$query = "SELECT * FROM $table_name WHERE (";
$countFields = count($arrayFields);
while ($a < $countFields)
{
while ($b < $countSearch)
{
$query = $query."$arrayFields[$a] LIKE '%$arraySearch[$b]%'";
$b++;
if ($b < $countSearch)
{
$query = $query." AND ";
}
}
$b = 0;
$a++;
if ($a < $countFields)
{
$query = $query.") OR (";
}
}
$query = $query.")";
$count_results = mysqli_query($connect, $query) or die(mysqli_error($connect));
$numrows = mysqli_num_rows($count_results);
// no results
if($numrows == 0) {
echo '<div id="counter">No results found</div>';
// show results
} else {
echo '<div id="results">
<div id="results_top"><p><b>'. $_GET['terms'] .'</b> - '. $numrows .' results found</p></div>
';
及
@CBroe帮我修好了!非常感谢他!问题在于AJAX和编码。我使用了
escape
函数代替encodeURIComponent
。这里是固定代码:
旧的:
http.open("GET", "search.php?terms=" + escape(zearch)+"&count="+get_count+"&page="+get_p, true);
固定的:
http.open("GET", "search.php?terms=" + encodeURIComponent(zearch)+"&count="+get_count+"&page="+get_p, true);
代码也是在线的。你可以在这里查看你在results div中对$\u GET['terms']的输出,是否正确显示?另外,您的AJAX代码在哪里?@CBroe我添加了AJAX代码,除Unicode字符外,其他代码都显示正确。您可以尝试键入“a”或“ä”查看结果。您发送的数据编码不正确,这会把结果搞砸。您不想使用
转义
,但要使用编码组件
。您正在请求,但它应该是(a
编码为%C3%A4
)
http.open("GET", "search.php?terms=" + escape(zearch)+"&count="+get_count+"&page="+get_p, true);
http.open("GET", "search.php?terms=" + encodeURIComponent(zearch)+"&count="+get_count+"&page="+get_p, true);