В настоящее время я работаю над PHP, и у меня есть 3 текстовых ввода. Значения выполняются в базе данных MySQL и должны возвращать любое количество результатов в соответствии с введенными критериями.
вот форма поиска:
<form id='SearchPersonal' method='post' action='businessUsersSearch.php' accept-charset='UTF-8'> <fieldset > <legend>Search</legend> <div class='container'> <label for='C_Name' >Business Name: </label><br/> <input type='text' name='C_Name' id='C_Name' maxlength="50" /><br/> <label for='C_County' >City: </label><br/> <input type='text' name='C_County' id='C_County' maxlength="50" /><br/> <label for='Job_Type' >Job Type: </label><br/> <input type='text' name='Job_Type' id='Job_Type' maxlength="50" /><br/> </div> <div class='container'> <input type='submit' name='Submit' value='Search' /> </div> </fieldset> </form> Вот скрипт PHP, который он также связывает в действии:
<?php $mysqli_link = mysqli_connect("server", "database", "pass", "user"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } if(isset($_POST['submit'])) { // define the list of fields $fields = array('C_Name', 'C_County', 'Job_Type'); $conditions = array(); // loop through the defined fields foreach($fields as $field){ // if the field is set and not empty if(isset($_POST[$field]) && $_POST[$field] != '') { // create a new condition while escaping the value inputed by the user (SQL Injection) $conditions[] = "'$field' LIKE '%" . mysqli_real_escape_string($mysqli_link, $_POST[$field]) . "%'"; } } // builds the query $query = "SELECT C_Name, C_StreetNumber, C_StreetName, C_Postcode, C_County, C_Tele, C_Website, Contact_Forename, Contact_Surname, Contact_Email, Jobs.Job_Type, Jobs.Job_Price FROM Company INNER JOIN Jobs ON Company.Company_ID = Jobs.Company_ID"; // if there are conditions defined if(count($conditions) > 0) { // append the conditions $query .= " WHERE " . implode (' AND ', $conditions); // you can change to 'OR', but I suggest to apply the filters cumulative } $result = mysqli_query($mysqli_link, $query) or die(mysql_error()); mysqli_close($mysqli_link); if(isset($_POST['submit'])) { while($row = mysqli_fetch_assoc($result)) { $C_Name = $row['C_Name']; $C_StreetNumber = $row['C_StreetNumber']; $C_StreetName = $row['C_StreetName']; $C_Postcode = $row['C_Postcode']; $C_County = $row['C_County']; $C_Tele = $row['C_Tele']; $C_Website = $row['C_Website']; $Contact_Forename = $row['Contact_Forename']; $Contact_Surname = $row['Contact_Surname']; $Contact_Email = $row['Contact_Email']; $Job_Type = $row['Job_Type']; $Job_Price = $row['Job_Price']; echo "<b>Name: $C_Name</b><br>Street Number: $C_StreetNumber<br>Street Name: $C_StreetName<br>Postcode: $C_Postcode<br>County: $C_County<br>Telephone: $C_Tele<br>Website: $C_Website<br>Contact Name: $Contact_Forename $Contact_Surname<br>Email: $Contact_Email<br>Job Type: $Job_Type<br>Job Price: $Job_Price<hr><br>"; } } } ?>
Почему-то возвращается, что есть "
неожиданный конец файла
«однако я проверил код, и все коды были закрыты правильно (из того, что я вижу), когда я добавляю еще один«} », в конце сценарий ничего не возвращает. Любой знает, почему это было бы происходит?
Источник: поиск базы данных MySQL с несколькими полями в форме
Потому что вы забыли закрыть
if(isset($_POST['submit'])) {// you not close the condition
В конце файла
Просто добавьте } в конец вашего файла
Исправлена:
if(isset($_POST['submit'])) { // define the list of fields $fields = array('C_Name', 'C_City', 'Job_Type', 'Review_Rate'); $conditions = array(); } // builds the query $query = "SELECT Company.C_Name, Company.C_StreetNumber, C_StreetName, C_Postcode, C_City, C_County, C_Tele, C_Website, Contact_Forename, Contact_Surname, Contact_Email, Job_Type, Job_Price, Review_Rate, Review_Comment FROM Company INNER JOIN Jobs ON Company.Company_ID = Jobs.Company_ID INNER JOIN Review ON Jobs.Job_ID = Review.Job_ID"; // loop through the defined fields foreach($fields as $field){ // if the field is set and not empty if(isset($_POST[$field]) && !empty($_POST[$field])) { // create a new condition while escaping the value inputed by the user (SQL Injection) $conditions[] = "$field LIKE '%" . mysqli_real_escape_string($mysqli_link, $_POST[$field]) . "%'"; } } // if there are conditions defined if(count($conditions) > 0) { // append the conditions $query .= " WHERE " . implode (' AND ', $conditions); // you can change to 'OR', but I suggest to apply the filters cumulative } echo "$query"; $result = mysqli_query($mysqli_link, $query); mysqli_close($mysqli_link); if(isset($_POST['submit'])) { while($row = mysqli_fetch_array($result)) { $C_Name = $row['C_Name']; $C_StreetNumber = $row['C_StreetNumber']; $C_StreetName = $row['C_StreetName']; $C_Postcode = $row['C_Postcode']; $C_City = $row['C_City']; $C_County = $row['C_County']; $C_Tele = $row['C_Tele']; $C_Website = $row['C_Website']; $Contact_Forename = $row['Contact_Forename']; $Contact_Surname = $row['Contact_Surname']; $Contact_Email = $row['Contact_Email']; $Job_Type = $row['Job_Type']; $Job_Price = $row['Job_Price']; $Rating = $row['Review_Rate']; $Comment = $row['Review_Comment']; echo "<b>Name: $C_Name</b><br>Street Number: $C_StreetNumber<br>Street Name: $C_StreetName<br>City: $C_City<br>Postcode: $C_Postcode<br>County: $C_County<br>Telephone: $C_Tele<br>Website: $C_Website<br>Contact Name: $Contact_Forename $Contact_Surname<br>Email: $Contact_Email<br>Job Type: $Job_Type<br>Job Price: $Job_Price<br>Rating: $Rating<br>Comment: $Comment<hr><br>"; } } ?>