Question

我试图在php中创建一个简单的搜索引擎，如果用户输入关键字，他们可以按事件名称或位置进行搜索。然后结果以日期顺序显示。大部分都来自另一个网站，但我试图转换它。

有人可以用简单的术语解释下面的错误给新手解释吗？

/home/ubuntu/workspace/test/test.php:49:array（5）{＆＃39; eventID＆＃39; =＆GT;字符串（1）＆＃34; 1＆＃34; ＆＃39; eventName的＆＃39; =＆GT; string（8）＆＃34;令人兴奋的＆＃34; ＆＃39; eventLocation＆＃39; =＆GT; string（7）＆＃34; Stadium＆＃34; ＆＃39;开始＆＃39; =＆GT; string（10）＆＃34; 2017-04-01＆＃34; ＆＃39;到期＆＃39; =＆GT; string（10）＆＃34; 2017-04-30＆＃34;警告：非法字符串偏移＆＃39; eventName＆＃39;在/home/ubuntu/workspace/test/test.php第50行调用堆栈：0.0003 238624 1. {main}（）/home/ubuntu/workspace/test/test.php:0 1

DisplayText

}

<?php
session_start();
//include files
include 'header/header.php';
include 'nav/navigation.php';
include 'init.php';
$expires = strtotime($_POST["expires"]);
$expires = date('Y-m-d', $expires);
$events =  "";
$find = $_POST['find'];
$field = $_POST['field'];
$searching = true;

if(isset($_POST["submit"])) {
if(!empty($_POST["events"])) {
    $searching = false;
}
   //This is only displayed if the user submitted the form 
if($searching == true) 
{
    echo "<h2>Results</h2><p>"; 
    //If the user did not enter a search term, they receive an error 
    if ($find == "") 
    { 
        echo "<p>You forgot to enter a search term"; 
        exit; 
    } 

    // Otherwise we connect to the database 
    //$result = mysqli_query($connection,$query) or exit ("Error in query: 
$query. ".mysqli_error($connection));

    // We preform a bit of filtering 
    $find = strtoupper($find); 
    $find = strip_tags($find); 
    $find = trim ($find); 

    //Now we search for our search term, in the field the user specified
    $mysqli = new mysqli('localhost', 'root', '', 'c3470438');
    $query = "SELECT * FROM events WHERE upper(eventLocation) 
LIKE'%STADIUM%'";
    $result = $mysqli->query($query);
    //Temporarily echo $query for debugging purposes    
    //echo "$query";
    //exit;
    //And display the results 
    $row = $result->fetch_array(MYSQLI_ASSOC);
    echo "<br>";
    foreach($row as $item) {
        var_dump($row);
        echo $item['eventName'];
        exit();
    }
    exit;


    //This counts the number or results. If there aren't any, it gives an 
explanation 
    $anymatches=mysql_num_rows($data); 
    if ($anymatches == 0) { 
        echo "Sorry, but we can not find an entry to match your query<br>
 <br>"; 
    } 

    //And reminds the user what they searched for 
    echo "<b>Searched For:</b> " .$find; 
}

Answer 1

好$result->fetch_array(MYSQLI_ASSOC)返回一行，但您正在使用它，就像它返回所有行一样。所以你的foreach迭代行中的每个字段，而不是每行，因此你试图访问字符串的数组键。

像这样使用一段时间：

while($row = $result->fetch_array(MYSQLI_ASSOC)){
    echo $row['eventName'];
}

此外，您使用的是不属于mysqli的mysql_num_rows。而是使用$result->num_rows。

警告：非法字符串偏移错误

1 个答案: