我使用PHP和jQuery创建了一个自动建议搜索框。提示用户在名为的表中插入名字和姓氏以查找数据库中存在的某人customers。表customers包含2列，first_name而last_name。
当您输入名字但按空格键移动并输入姓氏后，我的搜索工作正常。按下空格键时似乎出现了整个问题。知道如何解决吗？
$(document).ready(function($){
$("#customers").autocomplete({
source: "fetch_customers.php?cc=<?php echo $agencyid; ?>",
minLength: 2,
select: function(event, ui) {
var code = ui.item.id;
if (code != '#') {
location.href = '/view-customer/' + code;
}
},
open: function(event, ui) {
$(".ui-autocomplete").css("z-index", 1000);
}
});
});
<?php
$countrycode1 = $_GET['cc'];
$term = trim(strip_tags($_GET['term']));
$term = preg_replace('/\s+/', ' ', $term);
$a_json = array();
$a_json_row = array();
$a_json_invalid = array(array("id" => "#", "value" => $term, "label" => "Only letters and digits are permitted..."));
$json_invalid = json_encode($a_json_invalid);
if ($data = $conn->query("SELECT id, first_name, last_name FROM customers WHERE agency_id='$countrycode1' AND (first_name LIKE '%$term%' OR last_name LIKE '%$term%') ORDER BY first_name , last_name"))
{
while($row = mysqli_fetch_array($data))
{
$firstname = htmlentities(stripslashes($row['first_name']));
$lastname = htmlentities(stripslashes($row['last_name']));
$code = htmlentities(stripslashes($row['id']));
$a_json_row["id"] = $code;
$a_json_row["value"] = $firstname.' '.$lastname;
$a_json_row["label"] = $firstname.' '.$lastname;
array_push($a_json, $a_json_row);
}
}
/* jQuery wants JSON data */
$json = json_encode($a_json);
print $json;
flush();
$conn->close();
像这样拆分$ term。
$splited_term = explode(" ",$term);
$first_term = $splited_term[0];
$last_term = (isset($splited_term[1]) && !empty($splited_term)) ? $splited_term[1] : null;
根据生成查询
$query = "SELECT id, first_name, last_name FROM customers WHERE agency_id='$countrycode1' AND ";
if(!empty($first_term)){
$query.= "(first_name LIKE '%$first_term%' OR last_name LIKE '%$first_term%'" ;
}
if(!empty($last_term)){
$query .= ((!empty($first_term)) ? " OR " : " ( " )." first_name LIKE '%$last_term%' OR last_name LIKE '%$last_term%')" ;
}
$query .= ((empty($last_term)) ? ")" : "")." ORDER BY first_name , last_name";
这将支持
因此，您的代码应该像这样。
<?php
$countrycode1 = $_GET['cc'];
$term = trim(strip_tags($_GET['term']));
$term = preg_replace('/\s+/', ' ', $term);
$splited_term = explode(" ",$term);
$first_term = $splited_term[0];
$last_term = (isset($splited_term[1]) && !empty($splited_term)) ? $splited_term[1] : null;
$a_json = array();
$a_json_row = array();
$a_json_invalid = array(array("id" => "#", "value" => $term, "label" => "Only letters and digits are permitted..."));
$json_invalid = json_encode($a_json_invalid);
$query = "SELECT id, first_name, last_name FROM customers WHERE agency_id='$countrycode1' AND ";
if(!empty($first_term)){
$query.= "(first_name LIKE '%$first_term%' OR last_name LIKE '%$first_term%'" ;
}
if(!empty($last_term)){
$query .= ((!empty($first_term)) ? " OR " : " ( " )." first_name LIKE '%$last_term%' OR last_name LIKE '%$last_term%')" ;
}
$query .= ((empty($last_term)) ? ")" : "")." ORDER BY first_name , last_name";
if ($data = $conn->query($query))
{
while($row = mysqli_fetch_array($data))
{
$firstname = htmlentities(stripslashes($row['first_name']));
$lastname = htmlentities(stripslashes($row['last_name']));
$code = htmlentities(stripslashes($row['id']));
$a_json_row["id"] = $code;
$a_json_row["value"] = $firstname.' '.$lastname;
$a_json_row["label"] = $firstname.' '.$lastname;
array_push($a_json, $a_json_row);
}
}
/* jQuery wants JSON data */
$json = json_encode($a_json);
print $json;
flush();
$conn->close();
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