mysql_fetch_array（）期望参数1成为资源问题 [英] mysql_fetch_array() expects parameter 1 to be resource problem

问题描述

可能存在重复：

警告：mysql_fetch_array（）期望参数1为资源，布尔值为”错误，而尝试创建一个PHP购物车

我不明白，我在代码中看到没有错误，但是有这个错误，请帮助：

mysql_fetch_array（）期望参数1成为资源问题

 <？php 
 
 $ con = mysql_connect（localhost，root，nitoryolai123 $％^）; 
 if（！$ con）
 {
 die（'Could not connect：'。mysql_error（））; 
} 
 
 mysql_select_db（school，$ con）; 
 $ result = mysql_query（SELECT * FROM student WHERE IDNO =。$ _ GET ['id']）; 
？> 
 
 
 <？php while（$ row = mysql_fetch_array（$ result））{？> 
< table class =aborder =0align =centercellpadding =0cellspacing =1bgcolor =＃D3D3D3> 
< tr> 
 
 
< td> 
< table border =0cellpadding =3cellspacing =1bgcolor => 
< tr> 
 
< td colspan =16height =25style =background：＃5C915C; color：white; border：white 1px solid; text-align：left>< strong> ;< font size =2>更新学生< / td> 
 
 
< tr> 
< td width =30height =35>< font size =2> * I D编号：< / td> 
< td width =30>< input name =idnumonkeypress =return isNumberKey（event）type =textmaxlength =5id ='numbers'/ value = <？php echo $ _GET ['id'];？>>< / td> 
< / tr> 
 
< tr> 
< td width =30height =35>< font size =2> *年份：< / td> 
< td width =30>< input name =yronkeypress =return isNumberKey（event）type =textmaxlength =5id ='numbers'/ value = <？php echo $ row [YEAR];？>>< / td> 
 
<？php}？> 
  

我只是试图加载表单中的数据，但我不知道为什么会出现这种错误出现。这可能是什么错误？

在解析 之后，您没有进行错误检查 mysql_query ：

  $ result = mysql_query（SELECT * FROM student WHERE IDNO =。$ _ GET ['id']）; 
 if（！$ result）{//添加此检查。 
 die（'Invalid query：'。mysql_error（））; 
 
 $ / code> 

如果 mysql_query 失败时，它返回 false ，一个布尔值值。当你传递给 mysql_fetch_array 函数（它需要一个 mysql结果对象）时，我们会得到这个错误。

Possible Duplicate:
“Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given” error while trying to create a php shopping cart

I don't get it, I see no mistakes in this code but there is this error, please help:
mysql_fetch_array() expects parameter 1 to be resource problem

<?php

      $con = mysql_connect("localhost","root","nitoryolai123$%^");
    if (!$con)
      {
      die('Could not connect: ' . mysql_error());
      }

    mysql_select_db("school", $con);
       $result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']);
    ?>     


                           <?php while ($row = mysql_fetch_array($result)) { ?>             
                                     <table class="a"  border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#D3D3D3">
    <tr>

    <form name="formcheck" method="get" action="updateact.php" onsubmit="return formCheck(this);">
    <td>
    <table  border="0" cellpadding="3" cellspacing="1" bgcolor="">
    <tr>

    <td  colspan="16" height="25"  style="background:#5C915C; color:white; border:white 1px solid; text-align: left"><strong><font size="2">Update Students</td>


    <tr>
    <td width="30" height="35"><font size="2">*I D Number:</td>
    <td width="30"><input  name="idnum" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $_GET['id']; ?>"></td>
    </tr>

    <tr>
    <td width="30" height="35"><font size="2">*Year:</td>
    <td width="30"><input  name="yr" onkeypress="return isNumberKey(event)" type="text" maxlength="5" id='numbers'/ value="<?php echo $row["YEAR"]; ?>"></td>

<?php } ?>

I'm just trying to load the data in the forms but I don't know why that error appears. What could possibly be the mistake in here?

解决方案

You are not doing error checking after the call to mysql_query:

$result = mysql_query("SELECT * FROM student WHERE IDNO=".$_GET['id']);
if (!$result) { // add this check.
    die('Invalid query: ' . mysql_error());
}

In case mysql_query fails, it returns false, a boolean value. When you pass this to mysql_fetch_array function (which expects a mysql result object) we get this error.

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